| Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
Exercise 7.5
Question 1
Divide 12 into two parts such that their product is 32.
Sol :Let the first number be ‘X’, so the other number will be ’(12–X)’.
∵ X (12 – X) = 32
12X – X2 = 32
X2 – 12X + 32 = 0
On factorising further,
X2 – 4X – 8X + 32 = 0
X(X – 4) – 8(X – 4) = 0
(X – 4)(X – 8) = 0
So, X = 4 or 8
∴ The numbers are 4 and 8.
Question 2
Two numbers differ by 3 and their product is 504. Find the numbers.
Sol :Let the first number be ‘X’, so the other number will be ’(X+3)’.
∵ X(X + 3) = 504
X2 + 3X– 504=0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-3 \pm \sqrt{9-4(1)(-504)}}{2(1)}$
$\frac{-3 \pm \sqrt{9+2016}}{2(1)}$
$\frac{-3 \pm \sqrt{2025}}{2}$
$\frac{-3 \pm 45}{2}$
X = –24 or 21.
∴ if the first number is – 24 , then the other number is – 21.
∴ if the first number is 21 , then the other number is 24.
∴ The numbers are not (21,24) & (– 21, – 24).
Question 3
Find two consecutive positive integers, the sum of whose squares is 365.
Sol :Let the first number be ‘X’, so the other number will be ’(X+1)’.
∵ X2+(X+1)2 = 365
X2 + X2 + 1 + 2X – 365 = 0
2X2 + 2X – 364 = 0
X2 + X – 182 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-1 \pm \sqrt{1-4(1)(-182)}}{2(1)}$
$\frac{-1 \pm \sqrt{1+728}}{2(1)}$
$\frac{-1 \pm \sqrt{729}}{2}$
$\frac{-1 \pm 27}{2}$
∴ X = –14 or X = 13.
∴ The numbers are 13 & 14.
Question 4
The difference of two numbers is 4. If the difference of their reciprocals is $\frac{4}{21}$, find the two numbers.
Sol :Let the first number be ‘X’, so the other number will be ’(X+4)’.
$\because \frac{1}{X}-\frac{1}{X+4}=\frac{4}{21}$
$\frac{X+4-X}{X^{2}+4X}=\frac{4}{21}$
On simplifying further,
X2 + 4X – 21 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-4 \pm \sqrt{16-4(1)(-21)}}{2(1)}$
$\frac{-4 \pm \sqrt{16+84}}{2(1)}$
$\frac{-4 \pm \sqrt{100}}{2}$
$\frac{-4 \pm 10}{2}$
∴ X = –7 or X = 3.
∴ The numbers are –7 & –3 or 3 & 7.
Let the first number be ‘X’, so the other number will be ’(18–X)’.
$\because \frac{1}{X}+\frac{1}{18-X}=\frac{1}{4}$
$\frac{18-X+X}{-X^{2}+18 X}=\frac{1}{4}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-4 \pm \sqrt{16-4(1)(-21)}}{2(1)}$
$\frac{-4 \pm \sqrt{16+84}}{2(1)}$
$\frac{-4 \pm \sqrt{100}}{2}$
$\frac{-4 \pm 10}{2}$
∴ X = –7 or X = 3.
∴ The numbers are –7 & –3 or 3 & 7.
Question 5
The sum of two numbers is 18 and the sum of their reciprocals is $\frac{1}{4}$. Find the numbers.
Sol :Let the first number be ‘X’, so the other number will be ’(18–X)’.
$\because \frac{1}{X}+\frac{1}{18-X}=\frac{1}{4}$
$\frac{18-X+X}{-X^{2}+18 X}=\frac{1}{4}$
On simplifying further,
–X2 + 18X – 72 = 0
X2 – 18X + 72 = 0
–X2 + 18X – 72 = 0
X2 – 18X + 72 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-18) \pm \sqrt{(-18)^{2}-4(1)(72)}}{2(1)}$
$\frac{18 \pm \sqrt{324-288}}{2(1)}$
$\frac{18 \pm \sqrt{36}}{2}$
$\frac{18 \pm 6}{2}$
X = 6 or X = 12
∴ The numbers are 6 & 12.
Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).
∵X2 + (X + 1)2 + (X + 2)2 = 50
X2 + X2 + 1 + 2X + X2 + 4 + 4X = 50
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-18) \pm \sqrt{(-18)^{2}-4(1)(72)}}{2(1)}$
$\frac{18 \pm \sqrt{324-288}}{2(1)}$
$\frac{18 \pm \sqrt{36}}{2}$
$\frac{18 \pm 6}{2}$
X = 6 or X = 12
∴ The numbers are 6 & 12.
Question 6
The sum of the squares of three consecutive positive integers is 50. Find the integers.
Sol :Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).
∵X2 + (X + 1)2 + (X + 2)2 = 50
X2 + X2 + 1 + 2X + X2 + 4 + 4X = 50
On simplifying further,
3X2 + 6X– 45 = 0
X2 + 2X – 15 = 0
3X2 + 6X– 45 = 0
X2 + 2X – 15 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-2 \pm \sqrt{4-4(1)(-15)}}{2(1)}$
$\frac{-2 \pm \sqrt{4+60}}{2(1)}$
$\frac{-2 \pm \sqrt{64}}{2}$
$\frac{-2 \pm 8}{2}$
X = – 5 or 3
∴ X = 3 (Only Positive values)
∴ X + 1 = 4
∴ X +2 = 5
∴ The numbers are 3, 4 & 5.
Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).
∵ X2 + (X + 1)(X + 2) = 154
X2 + X2 + 2X + X + 2 = 154
On simplifying further,
2X2 + 3X – 152 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-2 \pm \sqrt{4-4(1)(-15)}}{2(1)}$
$\frac{-2 \pm \sqrt{4+60}}{2(1)}$
$\frac{-2 \pm \sqrt{64}}{2}$
$\frac{-2 \pm 8}{2}$
X = – 5 or 3
∴ X = 3 (Only Positive values)
∴ X + 1 = 4
∴ X +2 = 5
∴ The numbers are 3, 4 & 5.
Question 7
Find three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154.
Sol :Let the first number be ‘X’, so the other numbers will be ’(X+1)’ & ‘(X+2).
∵ X2 + (X + 1)(X + 2) = 154
X2 + X2 + 2X + X + 2 = 154
On simplifying further,
2X2 + 3X – 152 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-3 \pm \sqrt{3^{2}-4(2)(-152)}}{2(2)}$
$\frac{-3 \pm \sqrt{9+1216}}{4}$
$\frac{-3 \pm \sqrt{1225}}{4}$
$\frac{-3 \pm 35}{4}$
X = – 9.5 or 8
∴ X = 8 (Only whole values)
∴ X +1 = 9
∴ X + 2 = 10
∴ The numbers are 8, 9 & 10.
Let the units digit be ‘Y’, so the tens digit will be X.
∴ the number is 10X + Y
X . Y =14 –––––– (i)
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-3 \pm \sqrt{3^{2}-4(2)(-152)}}{2(2)}$
$\frac{-3 \pm \sqrt{9+1216}}{4}$
$\frac{-3 \pm \sqrt{1225}}{4}$
$\frac{-3 \pm 35}{4}$
X = – 9.5 or 8
∴ X = 8 (Only whole values)
∴ X +1 = 9
∴ X + 2 = 10
∴ The numbers are 8, 9 & 10.
Question 8
A two–digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
Sol :Let the units digit be ‘Y’, so the tens digit will be X.
∴ the number is 10X + Y
X . Y =14 –––––– (i)10X+Y+45=10Y+X
On simplifying further,
9X – 9Y + 45 = 0
X – Y + 5 = 0
On simplifying further,
9X – 9Y + 45 = 0
X – Y + 5 = 0
Putting the value of $\mathrm{X}=\frac{14}{\mathrm{Y}}$ from equation –––––– (i)
$\frac{14}{\mathrm{Y}}-\mathrm{Y}+5=0$
14 – Y2 + 5Y = 0
Y2 – 5Y – 14 = 0
$\frac{14}{\mathrm{Y}}-\mathrm{Y}+5=0$
14 – Y2 + 5Y = 0
Y2 – 5Y – 14 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-14)}}{2(1)}$
$\frac{5 \pm \sqrt{25+56}}{2}$
$\frac{5 \pm \sqrt{81}}{2}$
Y = – 2 or 7
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-14)}}{2(1)}$
$\frac{5 \pm \sqrt{25+56}}{2}$
$\frac{5 \pm \sqrt{81}}{2}$
$\frac{5 \pm 9}{2}$
Y = – 2 or 7
∴ Y = 7 (Only Positive values)
$\therefore \mathrm{X}=\frac{14}{7}$
∴ X = 2
∴ The number is 27 {2 (10) + 7}.
Let the larger number be ‘X’, so the smaller number will be ‘Y’.
∴ X2 – Y2 = 45 ––––– (i)
∵4X = Y2 –––––– (ii)
On simplifying further,
Putting the value of Y2 = 4X from equation –––––– (ii)
X2 – 4X = 45
X2 – 4X – 45 = 0
$\therefore \mathrm{X}=\frac{14}{7}$
∴ X = 2
∴ The number is 27 {2 (10) + 7}.
Question 9
The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Sol :Let the larger number be ‘X’, so the smaller number will be ‘Y’.
∴ X2 – Y2 = 45 ––––– (i)
∵4X = Y2 –––––– (ii)
On simplifying further,
Putting the value of Y2 = 4X from equation –––––– (ii)
X2 – 4X = 45
X2 – 4X – 45 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-45)}}{2(1)}$
$\frac{4 \pm \sqrt{16+180}}{2}$
$\frac{4 \pm \sqrt{196}}{2}$
$\frac{4 \pm 14}{2}$
X = – 5 or 9
∴ X = 9 (Only natural number, as given in the question)
∴ Y=√(4*9)
∴ Y = 6
∴ The numbers are 9 & 6.
Let the first number be ‘X’, so the other number will be ’(X+5)’.
$\because \frac{1}{X}-\frac{1}{X+5}=\frac{1}{10}$
$\frac{X+5-X}{X^{2}+5 X}=\frac{1}{10}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-4) \pm \sqrt{(-4)^{2}-4(1)(-45)}}{2(1)}$
$\frac{4 \pm \sqrt{16+180}}{2}$
$\frac{4 \pm \sqrt{196}}{2}$
$\frac{4 \pm 14}{2}$
X = – 5 or 9
∴ X = 9 (Only natural number, as given in the question)
∴ Y=√(4*9)
∴ Y = 6
∴ The numbers are 9 & 6.
Question 10
The difference of two numbers is 5 and the difference of their reciprocals is $\frac{1}{10}$Find the numbers.
Sol :Let the first number be ‘X’, so the other number will be ’(X+5)’.
$\because \frac{1}{X}-\frac{1}{X+5}=\frac{1}{10}$
$\frac{X+5-X}{X^{2}+5 X}=\frac{1}{10}$
On simplifying further,
X2 + 5X – 50 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-5 \pm \sqrt{5^{2}-4(1)(-50)}}{2(1)}$
$\frac{-5 \pm \sqrt{25+200}}{2(1)}$
$\frac{-5 \pm \sqrt{225}}{2}$
$\frac{-5 \pm 15}{2}$
∴ X = –10 or X = 5.
Then other numbers will be (X+5) = {–10+5} & {5+5} i.e., –5 or 10.
∴ The numbers are –10 & –5 or 5 & 10.
Let the number be ‘X’, so the reciprocal will be ’$\left(\frac{1}{x}\right)$’.
$\because \frac{1}{X}+X=\frac{10}{3}$
$\frac{1+X^{2}}{X}=\frac{10}{3}$
X2 + 5X – 50 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-5 \pm \sqrt{5^{2}-4(1)(-50)}}{2(1)}$
$\frac{-5 \pm \sqrt{25+200}}{2(1)}$
$\frac{-5 \pm \sqrt{225}}{2}$
$\frac{-5 \pm 15}{2}$
∴ X = –10 or X = 5.
Then other numbers will be (X+5) = {–10+5} & {5+5} i.e., –5 or 10.
∴ The numbers are –10 & –5 or 5 & 10.
Question 11
The sum of a number and its reciprocal is $\frac{10}{3}$. Find the number.
Sol :Let the number be ‘X’, so the reciprocal will be ’$\left(\frac{1}{x}\right)$’.
$\because \frac{1}{X}+X=\frac{10}{3}$
$\frac{1+X^{2}}{X}=\frac{10}{3}$
On simplifying further,
3 + 3X2 = 10X
3X2 – 10X + 3 = 0
3 + 3X2 = 10X
3X2 – 10X + 3 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-10) \pm \sqrt{(-10)^{2}-4(3)(3)}}{2(3)}$
$\frac{10 \pm \sqrt{100-36}}{6}$
$\frac{10 \pm \sqrt{64}}{6}$
$\frac{10 \pm 8}{6}$
∴ X = 3 or $X=\frac{1}{3}$
Then other numbers will be $\frac{1}{X}=\frac{1}{3}$& 3
∴ The numbers are 3 or $\frac{1}{3}$
Let the first number be ‘X’, so the other number will be ’(12–X)’.
∵ X2 + (12 – X)2 = 74
X2 + 144 + X2 – 24X = 74
On simplifying further,
2X2 – 24X + 70 = 0
X2 – 12X + 35 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-10) \pm \sqrt{(-10)^{2}-4(3)(3)}}{2(3)}$
$\frac{10 \pm \sqrt{100-36}}{6}$
$\frac{10 \pm \sqrt{64}}{6}$
$\frac{10 \pm 8}{6}$
∴ X = 3 or $X=\frac{1}{3}$
Then other numbers will be $\frac{1}{X}=\frac{1}{3}$& 3
∴ The numbers are 3 or $\frac{1}{3}$
Question 12
Divide 12 into two parts such that the sum of their squares is 74.
Sol :Let the first number be ‘X’, so the other number will be ’(12–X)’.
∵ X2 + (12 – X)2 = 74
X2 + 144 + X2 – 24X = 74
On simplifying further,
2X2 – 24X + 70 = 0
X2 – 12X + 35 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-12) \pm \sqrt{(-12)^{2}-4(1)(35)}}{2(1)}$
$\frac{12 \pm \sqrt{144-140}}{2(1)}$
$\frac{12 \pm \sqrt{4}}{2}$
$\frac{12 \pm 2}{2}$
∴ X = 5 or X = 7.
Then other numbers will be (12–X) = {12–5} & {12–7} i.e., 7 or 5.
∴ The number 12 is divided into two parts namely 5 & 7.
Let the first number be ‘X’, so the other number will be ’(X+1)’.
∵ X2 + (X + 1)2 = 421
X2 + X2 + 1 + 2X = 421
On simplifying further,
2X2+ 2X – 420 = 0
X2 + X – 210 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-12) \pm \sqrt{(-12)^{2}-4(1)(35)}}{2(1)}$
$\frac{12 \pm \sqrt{144-140}}{2(1)}$
$\frac{12 \pm \sqrt{4}}{2}$
$\frac{12 \pm 2}{2}$
∴ X = 5 or X = 7.
Then other numbers will be (12–X) = {12–5} & {12–7} i.e., 7 or 5.
∴ The number 12 is divided into two parts namely 5 & 7.
Question 13
The sum of the squares of two consecutive natural numbers is 421. Find the numbers.
Sol :Let the first number be ‘X’, so the other number will be ’(X+1)’.
∵ X2 + (X + 1)2 = 421
X2 + X2 + 1 + 2X = 421
On simplifying further,
2X2+ 2X – 420 = 0
X2 + X – 210 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(1) \pm \sqrt{(1)^{2}-4(1)(-210)}}{2(1)}$
$\frac{-1 \pm \sqrt{1+840}}{2(1)}$
$\frac{-1 \pm \sqrt{841}}{2}$
$\frac{-1 \pm 29}{2}$
∴ X = –15 or X = 14.
∴ X = 14 (Only natural number, as given in the question)
∴ Then other numbers will be 14 & 15.
Let the units digit be ‘Y’, so the tens digit will be X.
∴ the number is 10X + Y
∵ X . Y =18 –––––– (i)
10X+Y–63=10Y+X
On simplifying further,
9X – 9Y – 63 = 0
X – Y – 7 = 0
Putting the value of $X=\frac{18}{Y}$ from equation –––––– (i)
$\frac{18}{\mathrm{Y}}-\mathrm{Y}-7=0$
18 – Y2 – 7Y = 0
Y2 + 7Y – 18 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(1) \pm \sqrt{(1)^{2}-4(1)(-210)}}{2(1)}$
$\frac{-1 \pm \sqrt{1+840}}{2(1)}$
$\frac{-1 \pm \sqrt{841}}{2}$
$\frac{-1 \pm 29}{2}$
∴ X = –15 or X = 14.
∴ X = 14 (Only natural number, as given in the question)
∴ Then other numbers will be 14 & 15.
Question 14
A two–digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Sol :Let the units digit be ‘Y’, so the tens digit will be X.
∴ the number is 10X + Y
∵ X . Y =18 –––––– (i)
10X+Y–63=10Y+X
On simplifying further,
9X – 9Y – 63 = 0
X – Y – 7 = 0
Putting the value of $X=\frac{18}{Y}$ from equation –––––– (i)
$\frac{18}{\mathrm{Y}}-\mathrm{Y}-7=0$
18 – Y2 – 7Y = 0
Y2 + 7Y – 18 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-7 \pm \sqrt{(7)^{2}-4(1)(-18)}}{2(1)}$
$\frac{-7 \pm \sqrt{49+72}}{2}$
$\frac{-7 \pm \sqrt{121}}{2}$
$\frac{-7 \pm 11}{2}$
Y = – 9 or 2
Y = 2 (Only Positive values)
$\therefore \mathrm{X}=\frac{18}{2}$
∴ X = 9
∴ The number is 92 {9(10)+2}.
Let the units digit be ‘Y’, so the tens digit will be X.
the number is 10X + Y
10X + Y = 5 (X + Y)
On simplifying further,
10X – 5X – 5Y + Y = 0
5X – 4Y = 0 –––––– (i)
10X + Y = 2XY + 5
10X– 2XY + Y – 5 = 0 –––––– (ii)
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-7 \pm \sqrt{(7)^{2}-4(1)(-18)}}{2(1)}$
$\frac{-7 \pm \sqrt{49+72}}{2}$
$\frac{-7 \pm \sqrt{121}}{2}$
$\frac{-7 \pm 11}{2}$
Y = – 9 or 2
Y = 2 (Only Positive values)$\therefore \mathrm{X}=\frac{18}{2}$
∴ X = 9
∴ The number is 92 {9(10)+2}.
Question 15
A two–digit number is 5 times the sum of its digits and is also equal to 5 more than twice the product of its digits. Find the number.
Sol :Let the units digit be ‘Y’, so the tens digit will be X.
the number is 10X + Y 10X + Y = 5 (X + Y)On simplifying further,
10X – 5X – 5Y + Y = 0
5X – 4Y = 0 –––––– (i)
10X + Y = 2XY + 510X– 2XY + Y – 5 = 0 –––––– (ii)
Putting the value of $X=\frac{4 Y}{5}$ from equation –––––– (i)
$8 Y-\frac{8 Y^{2}}{5}+Y-5=0$
40Y – 8Y2 + 5Y – 25 = 0
8Y2 – 45Y + 25 = 0
$8 Y-\frac{8 Y^{2}}{5}+Y-5=0$
40Y – 8Y2 + 5Y – 25 = 0
8Y2 – 45Y + 25 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-45) \pm \sqrt{(-45)^{2}-4(8)(25)}}{2(8)}$
$\frac{45 \pm \sqrt{2025-800}}{16}$
$\frac{45 \pm \sqrt{1225}}{16}$
$\frac{45 \pm 35}{16}$
Y = $\frac{5}{8}$ or 5
∴ Y = 5
∴ $X=\frac{4 * 5}{5}$ from equation (i) $X=\frac{4 Y}{5}$
∴ X = 4
∴ The number is 45 {4(10)+5}.
Let the numerator be ‘X’, so the denominator will be ‘(2X+1)’.
the fraction is $\frac{X}{2 X+1}$
$\because \frac{X}{2 X+1}+\frac{2 X+1}{X}=\frac{58}{21}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-45) \pm \sqrt{(-45)^{2}-4(8)(25)}}{2(8)}$
$\frac{45 \pm \sqrt{2025-800}}{16}$
$\frac{45 \pm \sqrt{1225}}{16}$
$\frac{45 \pm 35}{16}$
Y = $\frac{5}{8}$ or 5
∴ Y = 5
∴ $X=\frac{4 * 5}{5}$ from equation (i) $X=\frac{4 Y}{5}$
∴ X = 4
∴ The number is 45 {4(10)+5}.
Question 16
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is $2 \frac{19}{15}$, find the fraction.
Sol :Let the numerator be ‘X’, so the denominator will be ‘(2X+1)’.
the fraction is $\frac{X}{2 X+1}$$\because \frac{X}{2 X+1}+\frac{2 X+1}{X}=\frac{58}{21}$
On simplifying further,
$\frac{X^{2}+4 X^{2}+4 X+1}{2 X^{2}+X}=\frac{58}{21}$
21(5X2 + 4X + 1) = 58 (2X2 + X)
105X2 + 84X + 21 = 116X2 + 58X
11X2 – 26X – 21 = 0–––––– (i)
$\frac{X^{2}+4 X^{2}+4 X+1}{2 X^{2}+X}=\frac{58}{21}$
21(5X2 + 4X + 1) = 58 (2X2 + X)
105X2 + 84X + 21 = 116X2 + 58X
11X2 – 26X – 21 = 0–––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-26) \pm \sqrt{(-26)^{2}-4(11)(-21)}}{2(11)}$
$\frac{26 \pm \sqrt{676+924}}{22}$
$\frac{26 \pm \sqrt{1600}}{22}$
$\frac{26 \pm 40}{22}$
X= $-\frac{7}{11}$ or 3
∴ X = 3 (Only positive values)
∴ Numerator is = 3 & denominator is (2X+1) = 7
∴ The fraction is $\frac{3}{7}$.
Given: Numerator of a fraction is one more than its denominator.
To Find: The fraction
Assumption: Let the denominator be x.
Numerator = x + 1
Therefore, the fraction $=\frac{x+1}{x}$
From the second case, we get,
$\frac{x+1}{x}-\frac{x}{x+1}=\frac{11}{30}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-26) \pm \sqrt{(-26)^{2}-4(11)(-21)}}{2(11)}$
$\frac{26 \pm \sqrt{676+924}}{22}$
$\frac{26 \pm \sqrt{1600}}{22}$
$\frac{26 \pm 40}{22}$
X= $-\frac{7}{11}$ or 3
∴ X = 3 (Only positive values)
∴ Numerator is = 3 & denominator is (2X+1) = 7
∴ The fraction is $\frac{3}{7}$.
Question 17
The numerator of a fraction is one more than its denominator. If its reciprocal is subtracted from it, the difference is $\frac{11}{30}$. Find the fraction.
Sol :Given: Numerator of a fraction is one more than its denominator.
To Find: The fraction
Assumption: Let the denominator be x.
Numerator = x + 1
Therefore, the fraction $=\frac{x+1}{x}$
From the second case, we get,
$\frac{x+1}{x}-\frac{x}{x+1}=\frac{11}{30}$
Taking L.C.M we get,
$\frac{(x+1)^{2}-x^{2}}{x(x+1)}=\frac{11}{30}$
$\frac{x^{2}+1+2 x-x^{2}}{x(x+1)}=\frac{11}{30}$
$\frac{2 x+1}{x^{2}+x}=\frac{11}{30}$
$\frac{(x+1)^{2}-x^{2}}{x(x+1)}=\frac{11}{30}$
$\frac{x^{2}+1+2 x-x^{2}}{x(x+1)}=\frac{11}{30}$
$\frac{2 x+1}{x^{2}+x}=\frac{11}{30}$
Cross-multiplying we get,
30(2x + 1) = 11(x2 + x)
60x + 30 = 11x2 + 11x
11x2 + 11x – 60x – 30 = 0
11x2 – 49x – 30 = 0
Now, we need to factorise such that, on multiplication we get 330 and on substraction we get 49.
Therefore, 55 and 6 can be the factors.
So, equation becomes,
11x2 – (55x – 6x) – 30 = 0
11x2 – 55x + 6x – 30 = 0
11x(x – 5) + 6(x – 5) = 0
(11x + 6)(x – 5) = 0
30(2x + 1) = 11(x2 + x)
60x + 30 = 11x2 + 11x
11x2 + 11x – 60x – 30 = 0
11x2 – 49x – 30 = 0
Now, we need to factorise such that, on multiplication we get 330 and on substraction we get 49.
Therefore, 55 and 6 can be the factors.
So, equation becomes,
11x2 – (55x – 6x) – 30 = 0
11x2 – 55x + 6x – 30 = 0
11x(x – 5) + 6(x – 5) = 0
(11x + 6)(x – 5) = 0
So, 11x + 5 = 0 or x – 5 = 0
$x=-\frac{5}{11}$ or $x=5$
$x+1=-\frac{5}{11}+1=\frac{6}{11}$ or
x+1=5+1=6
So the possible fractions are:
So the possible fractions are:
$\frac{x+1}{x}=\frac{\frac{6}{11}}{-\frac{5}{11}}=-\frac{6}{5}$
Or
$\frac{x+1}{x}=\frac{6}{5}$
Let the denominator be ‘X’, so the numerator will be ‘(X+1)’.
∴ the fraction is $\frac{\mathrm{XX}}{\mathrm{XX}+1}$
$\because \frac{X}{X+1}+\frac{X+1}{X}=\frac{61}{30}$
Or
$\frac{x+1}{x}=\frac{6}{5}$
Question 18
The numerator of a fraction is one more than its denominator. If its reciprocal is added to it the sum is $\frac{61}{30}$. Find the fraction.
Sol :Let the denominator be ‘X’, so the numerator will be ‘(X+1)’.
∴ the fraction is $\frac{\mathrm{XX}}{\mathrm{XX}+1}$
$\because \frac{X}{X+1}+\frac{X+1}{X}=\frac{61}{30}$
On simplifying further,
$\frac{X^{2}+X^{2}+2 X+1}{X^{2}+X}=\frac{61}{30}$
30(2X2 + 2X + 1) = 61 (X2 + X)
60X2 + 60X + 30 = 61X2+ 61X
X2 + X – 30 = 0–––––– (i)
$\frac{X^{2}+X^{2}+2 X+1}{X^{2}+X}=\frac{61}{30}$
30(2X2 + 2X + 1) = 61 (X2 + X)
60X2 + 60X + 30 = 61X2+ 61X
X2 + X – 30 = 0–––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-1 \pm \sqrt{(1)^{2}-4(1)(-30)}}{2(1)}$
$\frac{-1 \pm \sqrt{1+120}}{2}$
$\frac{-1 \pm \sqrt{121}}{2}$
$\frac{-1 \pm 11}{2}$
X= – 6 or 5
∴ X = 5 (Only positive values)
∴ Denominator is = 5 & numerator is (X+1) = 6
∴ The fraction is $\frac{6}{5}$.
Question 19
The numerator of a fraction is 3 more than its denominator. If its reciprocal is subtracted from it, the difference is $\frac{33}{28}$. Find the fraction.
Sol :Let the denominator be ‘X’, so the numerator will be ‘(X+3)’.
∴ the fraction is $\frac{X+3}{X}$
$\because \frac{X+3}{X}-\frac{X}{X+3}=\frac{33}{28}$
On simplifying further,
$\frac{X^{2}+6 X+9-X^{2}}{X^{2}+3 X}=\frac{33}{28}$
28(6X + 9) = 33(X2 + 3X)
168X + 252 = 33X2 + 99X
33X2 – 69X – 252 = 0
11X2 – 13X – 84 = 0 –––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-13) \pm \sqrt{(-13)^{2}-4(11)(-84)}}{2(11)}$
$\frac{13 \pm \sqrt{169+3696}}{22}$
$\frac{13 \pm \sqrt{3865}}{22}$
it does not have any real values.
Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+1}{X+4}$
$\because \frac{X+1}{X+4}-\frac{X}{X+3}=\frac{1}{24}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-13) \pm \sqrt{(-13)^{2}-4(11)(-84)}}{2(11)}$
$\frac{13 \pm \sqrt{169+3696}}{22}$
$\frac{13 \pm \sqrt{3865}}{22}$
it does not have any real values.Question 20
The denominator of a fraction exceeds its numerator by 3. If one is added to both numerator and denominator, the difference between the new and the original fractions 1 becomes $\frac{1}{24}$. Find the original fraction.
Sol :Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+1}{X+4}$
$\because \frac{X+1}{X+4}-\frac{X}{X+3}=\frac{1}{24}$
On simplifying further,
$\frac{X^{2}+3 X+X+3-X^{2}-4 X}{X^{2}+3 X+4 X+12}=\frac{1}{24}$
24(3) = X2 + 7X + 12
X2 + 7X – 60 = 0 –––––– (i)
$\frac{X^{2}+3 X+X+3-X^{2}-4 X}{X^{2}+3 X+4 X+12}=\frac{1}{24}$
24(3) = X2 + 7X + 12
X2 + 7X – 60 = 0 –––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(7) \pm \sqrt{(7)^{2}-4(1)(-60)}}{2(1)}$
$\frac{-7 \pm \sqrt{49+240}}{2}$
$\frac{-7 \pm \sqrt{289}}{2}$
$\frac{-7 \pm 17}{2}$
∴ X = –12 or 5.
∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\frac{5}{8}$, as taking (–12) will form a fraction i.e., $\frac{4}{3}$ (not satisfying the conditions) .
Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+3}{X+6}$
$\because \frac{X+3}{X+6}-\frac{X}{X+3}=\frac{9}{88}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(7) \pm \sqrt{(7)^{2}-4(1)(-60)}}{2(1)}$
$\frac{-7 \pm \sqrt{49+240}}{2}$
$\frac{-7 \pm \sqrt{289}}{2}$
$\frac{-7 \pm 17}{2}$
∴ X = –12 or 5.
∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\frac{5}{8}$, as taking (–12) will form a fraction i.e., $\frac{4}{3}$ (not satisfying the conditions) .
Question 21
The denominator of a fraction exceeds its numerator by 3. If 3 is added to both numerator and denominator, the difference between the new and the original fraction is $\frac{9}{88}$. Find the original fraction.
Sol :Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+3}{X+6}$
$\because \frac{X+3}{X+6}-\frac{X}{X+3}=\frac{9}{88}$
On simplifying further,
$\frac{X^{2}+6 X+9-X^{2}-6 X}{X^{2}+3 X+6 X+18}=\frac{9}{88}$
88(9) = 9(X2 + 9X + 18)
X2 + 9X – 70 = 0 –––––– (i)
$\frac{X^{2}+6 X+9-X^{2}-6 X}{X^{2}+3 X+6 X+18}=\frac{9}{88}$
88(9) = 9(X2 + 9X + 18)
X2 + 9X – 70 = 0 –––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(9) \pm \sqrt{(9)^{2}-4(1)(-70)}}{2(1)}$
$\frac{-9 \pm \sqrt{81+280}}{2}$
$\frac{-9 \pm \sqrt{361}}{2}$
$\frac{-9 \pm 19}{2}$
∴ X = –14 or 5.
∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\frac{5}{8}$, as taking (–14) will form a fraction i.e., $\frac{14}{11}$ (not satisfying the conditions) .
Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+2}{X+5}$
$\because \frac{X+2}{X+5}+\frac{X}{X+3}=\frac{19}{15}$
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(9) \pm \sqrt{(9)^{2}-4(1)(-70)}}{2(1)}$
$\frac{-9 \pm \sqrt{81+280}}{2}$
$\frac{-9 \pm \sqrt{361}}{2}$
$\frac{-9 \pm 19}{2}$
∴ X = –14 or 5.
∴ the numerator is X = 5 and denominator (X+3) will be 8 and the fraction will be $\frac{5}{8}$, as taking (–14) will form a fraction i.e., $\frac{14}{11}$ (not satisfying the conditions) .
Question 22
The numerator of a fraction is 3 less than denominator. If 2 is added to both 29 numerator as well as denominator, then sum of the new and original fraction is $\frac{19}{15}$ . Find the fraction.
Sol :Let the denominator be ‘(X+3)’, so the numerator will be ‘X’.
∴ the original fraction is $\frac{X}{X+3}$
∴ the new fraction is $\frac{X+2}{X+5}$
$\because \frac{X+2}{X+5}+\frac{X}{X+3}=\frac{19}{15}$
On simplifying further,
$\frac{X^{2}+5 X+6+X^{2}+5 X}{X^{2}+3 X+5 X+15}=\frac{19}{15}$
15(2X2 + 10X + 6) = 19(X2 + 8X + 15)
30X2 + 150X + 90 = 19X2 + 152X + 285
11X2 – 2X – 195 = 0 –––––– (i)
$\frac{X^{2}+5 X+6+X^{2}+5 X}{X^{2}+3 X+5 X+15}=\frac{19}{15}$
15(2X2 + 10X + 6) = 19(X2 + 8X + 15)
30X2 + 150X + 90 = 19X2 + 152X + 285
11X2 – 2X – 195 = 0 –––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-2) \pm \sqrt{(-2)^{2}-4(11)(-195)}}{2(11)}$
$\frac{2 \pm \sqrt{4+8580}}{22}$
$\frac{2 \pm \sqrt{8584}}{22}$
∴ it does not have real values.
Given: Numerator of a fraction is 2 less than the denominator
To find: The fraction
Assumption: Let the denominator be x
Numerator = x – 2
Therefore, the fraction $=\frac{x-2}{x}$
If one is added to the numerator and denominator, fraction becomes
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-2) \pm \sqrt{(-2)^{2}-4(11)(-195)}}{2(11)}$
$\frac{2 \pm \sqrt{4+8580}}{22}$
$\frac{2 \pm \sqrt{8584}}{22}$
∴ it does not have real values.
Question 23
The numerator of a fraction is 2 less than the denominator. If 1 is added to both numerator and denominator the sum of the new and original fraction is $\frac{19}{15}$. Find the original fraction.
Sol :Given: Numerator of a fraction is 2 less than the denominator
To find: The fraction
Assumption: Let the denominator be x
Numerator = x – 2
Therefore, the fraction $=\frac{x-2}{x}$
If one is added to the numerator and denominator, fraction becomes
$=\frac{x-2+1}{x+1}=\frac{x-1}{x+1}$
Sum of the fractions $=\frac{x-2}{x}+\frac{x-1}{x+1}$
Sum of the fractions $=\frac{19}{15}$
Therefore,
$\frac{x-2}{x}+\frac{x-1}{x+1}=\frac{19}{15}$
Taking L.C.M we get,
$\frac{(x-2)(x+1)+x(x-1)}{x(x+1)}=\frac{19}{15}$
$\frac{x^{2}+x-2 x-2+x^{2}-x}{x^{2}+x}=\frac{19}{15}$
$\frac{2 x^{2}-2 x-2}{x^{2}+x}=\frac{19}{15}$
Cross-multiplying we get,
30x2 – 30x – 30 = 19x2 + 19x
30x2 – 19x2 – 30x – 19x – 30 = 0
11x2 – 49x – 30 = 0
Now we need to factorise such that, on multiplication we get 330 and on substraction we get 49.
So, equation becomes,
11x2 – (55x – 6x) – 30 = 0
11x2 – 55x + 6x – 30 = 0
11x(x – 5) + 6(x – 5) = 0
(11x + 6)(x – 5) = 0
So, 11x + 5 = 0 or x – 5 = 0
$x=-\frac{5}{11}$ or x=5
$x-2=-\frac{5}{11}-2=-\frac{27}{11}$ or
$x-2=-\frac{5}{11}-2=-\frac{27}{11}$ or
x=5-2=3
Putting these values in fraction $=\frac{x-2}{x}$
Hence, the possible fractions are,
$\frac{x-2}{x}=\frac{-\frac{27}{11}}{-\frac{5}{11}}=\frac{27}{5}$
$\frac{x-2}{x}=\frac{3}{5}$
Let the shortest side(AC) be ‘(X)’cms, so the hypotenuse (BC) will be ‘(2X+6)’ cms, as demonstrated in the figure drawn below:
∴ AB = 2X + 4 cms
∵ (BC)2 = (AC)2 + (AB)2
∴(2X + 6)2 = X2 +(2X + 4)2
On simplifying further,
4X2 + 24X + 36 = X2 + 4X2 + 16X + 16
Using the identity of a2 + b2 + 2ab = (a + b)2
X2 – 8X – 20 = 0 –––––– (i)
Putting these values in fraction $=\frac{x-2}{x}$
Hence, the possible fractions are,
$\frac{x-2}{x}=\frac{-\frac{27}{11}}{-\frac{5}{11}}=\frac{27}{5}$
$\frac{x-2}{x}=\frac{3}{5}$
Question 24
The hypotenuse of a right–angled triangle is 6 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.
Sol :Let the shortest side(AC) be ‘(X)’cms, so the hypotenuse (BC) will be ‘(2X+6)’ cms, as demonstrated in the figure drawn below:
∴ AB = 2X + 4 cms
∵ (BC)2 = (AC)2 + (AB)2
∴(2X + 6)2 = X2 +(2X + 4)2
On simplifying further,
4X2 + 24X + 36 = X2 + 4X2 + 16X + 16
Using the identity of a2 + b2 + 2ab = (a + b)2
X2 – 8X – 20 = 0 –––––– (i)
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(-20)}}{2(1)}$
$\frac{8 \pm \sqrt{64+80}}{2}$
$\frac{8 \pm \sqrt{144}}{2}$
$\frac{8 \pm 12}{2}$
∴ X = –2 or 10.
∴ the shortest side (AC) is X = 10 cms (Only positive values), hypotenuse (2X+6) i.e., (BC) is 26 cms and the other side (AB) is (2X+4) i.e., 24 cms .
Let the sides of the squares be ‘(X)’cms and ‘(Y) cms.
∴ X2 + Y2 = 640 –––––(i)
∵ Area = (Side)2
∵Perimeter = 4 (Side)
∴4X – 4Y = 64
On simplifying further,
X – Y = 16 ––––––(ii)
Squaring the above mentioned equation, i.e., equation (ii)
X2 + Y2 – 2XY = 256
Using the identity of a2 + b2 – 2ab = (a – b)2
Putting the value of equation (i) in equation (ii)
∴ 640 – 2XY = 256
2XY = 384
XY = 192
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-8) \pm \sqrt{(-8)^{2}-4(1)(-20)}}{2(1)}$
$\frac{8 \pm \sqrt{64+80}}{2}$
$\frac{8 \pm \sqrt{144}}{2}$
$\frac{8 \pm 12}{2}$
∴ X = –2 or 10.
∴ the shortest side (AC) is X = 10 cms (Only positive values), hypotenuse (2X+6) i.e., (BC) is 26 cms and the other side (AB) is (2X+4) i.e., 24 cms .
Question 25
The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.
Sol :Let the sides of the squares be ‘(X)’cms and ‘(Y) cms.
∴ X2 + Y2 = 640 –––––(i)
∵ Area = (Side)2
∵Perimeter = 4 (Side)
∴4X – 4Y = 64
On simplifying further,
X – Y = 16 ––––––(ii)
Squaring the above mentioned equation, i.e., equation (ii)
X2 + Y2 – 2XY = 256
Using the identity of a2 + b2 – 2ab = (a – b)2
Putting the value of equation (i) in equation (ii)
∴ 640 – 2XY = 256
2XY = 384
XY = 192
Putting the value of X+Y=16 in equation (i)
(Y+16)2 + Y2 = 640
Y2 + 256 + 32Y + Y2 = 640
2Y2 + 32Y – 384 = 0
Y2 + 16Y – 192 = 0
(Y+16)2 + Y2 = 640
Y2 + 256 + 32Y + Y2 = 640
2Y2 + 32Y – 384 = 0
Y2 + 16Y – 192 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(16) \pm \sqrt{(16)^{2}-4(1)(-192)}}{2(1)}$
$\frac{-16 \pm \sqrt{256+768}}{2}$
$\frac{-16 \pm \sqrt{1024}}{2}$
$\frac{-16 \pm 32}{2}$
∴ X = – 12 or 24.
∴ the side is X = 24 cms (Only positive values), other square’s side is Y=X–16 i.e., 8 cms.
Let the shortest side(AC) be ‘(X)’cms, and the longer side (AB) be ‘(Y)’ cms, as demonstrated in the figure drawn below:
∴ BC = 3√10 cms
∵ (BC)2 = (AC)2 + (AB)2
∴ (3√10)2 = X2 + (Y)2
On simplifying further,
(Y)2 = 90 – X2
X2 + Y2 – 90 = 0 –––––– (i)
As per the question,
New smaller side = ‘(3X)’ cms
New longer side = ‘(2Y)’ cms
∴ BC = 9√5 cms
∵ (BC)2 = (AC)2 + (AB)2
∴(9√5)2 = (3X)2 + (2Y)2
On simplifying further,
4Y2 + 9X2 = 405
4X2 + 9Y2 – 405 = 0 –––––– (ii)
Putting the value of X2, from equation (i) in equation (ii)
X2 = 90 – Y2
On simplifying further,
(360–4Y)2 + 9Y2 – 405 = 0
5Y2 = 45
Y = ± 3 cms(Only positive values),
∴ X = √(90–9).
∴ X = 9 cms
∴ the shortest side (AC) is X = 9 cms hypotenuse i.e., (BC) is $3 \sqrt{10}$ cms and the other side (AB) is 3 cms .
Ok now we know that area of square = side2
So no. of students in the line if side was x (assumption)
No. of students = x2+24 (as there were 24 exrltra students)
No. Of students after increasing 1 student in square = (x+1)2–25 (as there were 25 less students)
So x2 + 24 = (x + 1)2 – 25
x2+ 24 = x2 + 1 + 2x – 25
x2 – x2 + 24 + 25 – 1 = 2x
48 = 2x
So x = 24
So no. Of students = x2 + 24 = 242 + 24 = 576 + 24 = 600
∴ the no. of students = 600.
Let the length of base = P cm.
As base exceeds the base by 7cm ,
then , length of altitude = (P + 7)cm
now, area of triangle $=\frac{1}{2} \times$ altitude $\times$ base
given, area of triangle = 30 cm2
so, $30 \mathrm{cm}^{2}=\frac{1}{2} \times(\mathrm{P}+7) \times \mathrm{P}$
⇒ 30 × 2 = P2 + 7P
⇒ 60 = P2 + 7P
⇒ P2 + 7P – 60 = 0
⇒ P2 + 12P – 5P – 60 = 0
⇒ P(P + 12) – 5(P + 12) = 0
⇒ (P + 12)(P –5) = 0
⇒ P = 5 , –12
but length can't be negative so, P ≠ –12
hence, P = 5 cm e.g., base = 5cm
Let breadth be X cm
length = 2X cms
area=800
2X (X)=800
X(X)=400
X2 = 400
X =20
Yes it is possible to design a rectangular mangrove having breadth =20 cm and length=40 cm.
Let the length be
then breadth= l – 3
area of rectangle= l(l – 3) = l2 – 3l
area of triangle$=\frac{1}{2}(1-3)(12)=61-18$
given
area of rectangle is 4sq mt more than triangle
so
area of rectangle – 4= area of triangle
l2 – 3l – 4 = 6l – 18
l2 – 9l + 14 = 0
l2 – 7l – 2l + 14 = 0
l(l – 7) – 2(l – 7) = 0
(l – 7)(l – 2) = 0
Yes, it is possible, to design a rectangular park.
L = 7 and 2
L = 2 is neglected as when length is 2 the breadth will be negative which is not possible...
so length= 7m
breadth= 1-3=7–3 = 4m
Yes it is possible to design a rectangular park having length & breadth 7 mts & 4 mts respectively.
Let P be the position of the pole and A and B be the opposite fixed gates.
PA = ‘a’ mts
PB = ‘b’ mts
PA – PB = 7 m
⇒ a – b = 7
⇒ a = 7 + b .........(1)
In Δ PAB,
AB2 = AP2 + BP2
⇒ (17) = (a)2 + (b)2
⇒ a2 + b2 = 289
⇒ Putting the value of a = 7 + b in the above,
(7 + b)2 + b2 = 289
⇒ 49 + 14b + 2b2 = 289
⇒ 2b2 + 14b + 49 – 289 = 0
⇒ 2b2 + 14b – 240 = 0
Dividing the above by 2, we get.
⇒ b2 + 7b – 120 = 0
⇒ b2 + 15b – 8b – 120 = 0
⇒ b(b + 15) – 8(b + 15) = 0
⇒ (b – 8) (b + 15) = 0
⇒ b = 8 or b = –15
Since this value cannot be negative, so b = 8 is the correct value.
Yes it is possible to erect a pole.
Putting b = 8 in (1), we get.
a = 7 + 8
a = 15 m
Hence PA = 15 m and PB = 8 m
So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m.
Let the mother age be ‘X’ years, then her daughter’s age will be ‘(20–X)’ years.
As per the question,
(X – 4)(20 – X – 4) = 48
(X – 4)(16 – X) = 48
16X – 64 – X2 + 4X = 48
–X2 + 20X – 112 = 0
X2 – 20X + 112 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-20) \pm \sqrt{(-20)^{2}-4(1)(112)}}{2(1)}$
$\frac{20 \pm \sqrt{400+448}}{2}$
$\frac{20 \pm \sqrt{848}}{2}$
∵ it does not have real values, then it is not possible for the above situation to happen.
Let the speed of the train and the time taken to cover the same be ‘X’ km/hr and ‘Y’ hrs respectively.
∵Distance=speed*time
∵ As per the question,
XY = 90 –––––– (i)
(X + 15)(Y – 0.5) = 90 ––––– (ii)
∵ LHS = RHS
∴ Equating the LHS of both equations, and simplyifying it further
XY = XY – 0.5X + 15Y – 7.5
0.5X – 15Y + 7.5 = 0 ––––––– (iii)
Multiplying the above equation by 10,
5X – 150Y + 75 = 0
Simplyfying it further,
X – 30Y + 15 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(16) \pm \sqrt{(16)^{2}-4(1)(-192)}}{2(1)}$
$\frac{-16 \pm \sqrt{256+768}}{2}$
$\frac{-16 \pm \sqrt{1024}}{2}$
$\frac{-16 \pm 32}{2}$
∴ X = – 12 or 24.
∴ the side is X = 24 cms (Only positive values), other square’s side is Y=X–16 i.e., 8 cms.
Question 26
The hypotenuse of a right triangle is 3√5cm. If the smaller side is tripled and the longer side doubled, new hypotenuse will be 9√5 cm. How long are the sides of the triangle?
Sol :Let the shortest side(AC) be ‘(X)’cms, and the longer side (AB) be ‘(Y)’ cms, as demonstrated in the figure drawn below:
∴ BC = 3√10 cms
∵ (BC)2 = (AC)2 + (AB)2
∴ (3√10)2 = X2 + (Y)2
On simplifying further,
(Y)2 = 90 – X2
X2 + Y2 – 90 = 0 –––––– (i)
As per the question,
New smaller side = ‘(3X)’ cms
New longer side = ‘(2Y)’ cms
∴ BC = 9√5 cms
∵ (BC)2 = (AC)2 + (AB)2
∴(9√5)2 = (3X)2 + (2Y)2
On simplifying further,
4Y2 + 9X2 = 405
4X2 + 9Y2 – 405 = 0 –––––– (ii)
Putting the value of X2, from equation (i) in equation (ii)
X2 = 90 – Y2
On simplifying further,
(360–4Y)2 + 9Y2 – 405 = 0
5Y2 = 45
Y = ± 3 cms(Only positive values),
∴ X = √(90–9).
∴ X = 9 cms
∴ the shortest side (AC) is X = 9 cms hypotenuse i.e., (BC) is $3 \sqrt{10}$ cms and the other side (AB) is 3 cms .
Question 27
A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that lie was short of 25 students. Find the number of students.
Sol :Ok now we know that area of square = side2
So no. of students in the line if side was x (assumption)
No. of students = x2+24 (as there were 24 exrltra students)
No. Of students after increasing 1 student in square = (x+1)2–25 (as there were 25 less students)
So x2 + 24 = (x + 1)2 – 25
x2+ 24 = x2 + 1 + 2x – 25
x2 – x2 + 24 + 25 – 1 = 2x
48 = 2x
So x = 24
So no. Of students = x2 + 24 = 242 + 24 = 576 + 24 = 600
∴ the no. of students = 600.
Question 28
The area of a triangle is 30 sq cm. Find the base if the altitude exceeds the base by 7 cm.
Sol :Let the length of base = P cm.
As base exceeds the base by 7cm ,
then , length of altitude = (P + 7)cm
now, area of triangle $=\frac{1}{2} \times$ altitude $\times$ base
given, area of triangle = 30 cm2
so, $30 \mathrm{cm}^{2}=\frac{1}{2} \times(\mathrm{P}+7) \times \mathrm{P}$
⇒ 30 × 2 = P2 + 7P
⇒ 60 = P2 + 7P
⇒ P2 + 7P – 60 = 0
⇒ P2 + 12P – 5P – 60 = 0
⇒ P(P + 12) – 5(P + 12) = 0
⇒ (P + 12)(P –5) = 0
⇒ P = 5 , –12
but length can't be negative so, P ≠ –12
hence, P = 5 cm e.g., base = 5cm
Question 29
Is it possible to design a rectangular mango grove whose length is twice its breadth, and area is 800 m2? If so, find its length and breadth.
Sol :Let breadth be X cm
length = 2X cms
area=800
2X (X)=800
X(X)=400
X2 = 400
X =20
Yes it is possible to design a rectangular mangrove having breadth =20 cm and length=40 cm.
Question 30
I want to design a rectangular park whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as breadth of the rectangular park and altitude 12 m. Is it possible to have such a rectangular park? If so, find its length and breadth.
Sol :Let the length be
then breadth= l – 3
area of rectangle= l(l – 3) = l2 – 3l
area of triangle$=\frac{1}{2}(1-3)(12)=61-18$
given
area of rectangle is 4sq mt more than triangle
so
area of rectangle – 4= area of triangle
l2 – 3l – 4 = 6l – 18
l2 – 9l + 14 = 0
l2 – 7l – 2l + 14 = 0
l(l – 7) – 2(l – 7) = 0
(l – 7)(l – 2) = 0
Yes, it is possible, to design a rectangular park.
L = 7 and 2
L = 2 is neglected as when length is 2 the breadth will be negative which is not possible...
so length= 7m
breadth= 1-3=7–3 = 4m
Yes it is possible to design a rectangular park having length & breadth 7 mts & 4 mts respectively.
Question 31
A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Sol :Let P be the position of the pole and A and B be the opposite fixed gates.
PA = ‘a’ mts
PB = ‘b’ mts
PA – PB = 7 m
⇒ a – b = 7
⇒ a = 7 + b .........(1)
In Δ PAB,
AB2 = AP2 + BP2
⇒ (17) = (a)2 + (b)2
⇒ a2 + b2 = 289
⇒ Putting the value of a = 7 + b in the above,
(7 + b)2 + b2 = 289⇒ 49 + 14b + 2b2 = 289
⇒ 2b2 + 14b + 49 – 289 = 0
⇒ 2b2 + 14b – 240 = 0
Dividing the above by 2, we get.
⇒ b2 + 7b – 120 = 0
⇒ b2 + 15b – 8b – 120 = 0
⇒ b(b + 15) – 8(b + 15) = 0
⇒ (b – 8) (b + 15) = 0
⇒ b = 8 or b = –15
Since this value cannot be negative, so b = 8 is the correct value.
Yes it is possible to erect a pole.
Putting b = 8 in (1), we get.
a = 7 + 8
a = 15 m
Hence PA = 15 m and PB = 8 m
So, the distance from the gate A to pole is 15 m and from gate B to the pole is 8 m.
Question 32
Is the following situation possible? If so, determine their present ages. The sum of the ages of a mother and her daughter is 20 years. Four years ago, the product of their ages in years was 48.
Sol :Let the mother age be ‘X’ years, then her daughter’s age will be ‘(20–X)’ years.
As per the question,
(X – 4)(20 – X – 4) = 48
(X – 4)(16 – X) = 48
16X – 64 – X2 + 4X = 48
–X2 + 20X – 112 = 0
X2 – 20X + 112 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-20) \pm \sqrt{(-20)^{2}-4(1)(112)}}{2(1)}$
$\frac{20 \pm \sqrt{400+448}}{2}$
$\frac{20 \pm \sqrt{848}}{2}$
∵ it does not have real values, then it is not possible for the above situation to happen.
Question 33
A train covers a distance of 90 km at a uniform speed. Had the speed been 15 kmph more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Sol :Let the speed of the train and the time taken to cover the same be ‘X’ km/hr and ‘Y’ hrs respectively.
∵Distance=speed*time
∵ As per the question,
XY = 90 –––––– (i)
(X + 15)(Y – 0.5) = 90 ––––– (ii)
∵ LHS = RHS
∴ Equating the LHS of both equations, and simplyifying it further
XY = XY – 0.5X + 15Y – 7.5
0.5X – 15Y + 7.5 = 0 ––––––– (iii)
Multiplying the above equation by 10,
5X – 150Y + 75 = 0
Simplyfying it further,
X – 30Y + 15 = 0
Putting the value of Y, in equation (iii)
$x-30\left(\frac{90}{x}\right)+15=0$
X2 + 15X – 2700 = 0
$x-30\left(\frac{90}{x}\right)+15=0$
X2 + 15X – 2700 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(15) \pm \sqrt{(15)^{2}-4(1)(2700)}}{2(1)}$
$\frac{-15 \pm \sqrt{225+10800}}{2}$
$\frac{-15 \pm \sqrt{11025}}{2}$
$\frac{-15 \pm 105}{2}$
X = – 60 or X = 45
∴ the original speed of train is 45 km/hr as speed can’t be negative.
Let the usual speed be x km /hr.
Actual speed = (x + 250) km/hr.
Time taken at actual speed = ($\frac{1500}{x}$) hr.
Difference between the two times taken $=\frac{1}{2}$ hr.
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(15) \pm \sqrt{(15)^{2}-4(1)(2700)}}{2(1)}$
$\frac{-15 \pm \sqrt{225+10800}}{2}$
$\frac{-15 \pm \sqrt{11025}}{2}$
$\frac{-15 \pm 105}{2}$
X = – 60 or X = 45
∴ the original speed of train is 45 km/hr as speed can’t be negative.
Question 34
An aeroplane left 30 minutes later than its scheduled time and in order to reach its destination 1500 km away in time, it had to increase its speed by 250 km/hr from its usual speed. Determine its usual speed.
Sol :Let the usual speed be x km /hr.
Actual speed = (x + 250) km/hr.
Time taken at actual speed = ($\frac{1500}{x}$) hr.
Difference between the two times taken $=\frac{1}{2}$ hr.
Speed at that time = (x + 250) km/hr
$\frac{\text { Distance }}{\text { time }}=\mathrm{x}+250$
$\frac{1500}{\text { time }}=x+250$
$\frac{1500}{x+250}=$ time
$\frac{1500}{\text { time }}=x+250$
$\frac{1500}{x+250}=$ time
Then, According to the question,
$\frac{1500}{x}-\frac{1500}{x+250}=30 \times \frac{1}{60}$
$1500\left(\frac{1}{x}-\frac{1}{(x+250)}\right)=\frac{1}{2}$
$3000 \times\left[\frac{x+250-x}{x^{2}+250}\right]=1$
3000{250} = x2 + 250
0 = x2 + 250x – 750000
0 = x2 + (1000–750)x – 750000
0 = x2 + 1000x – 750x – 750000
0 = x(x + 1000) – 750(x + 1000)
0 = (x + 1000) (x – 750)
x = –1000 or x = 750
Usual speed = x = 750 km /hr
⇒ x = 750 【 speed cannot be negative】
Hence , the usual speed of the aeroplane was 750 km / hr.
Question 35
The speed of a boat in still water is 15 km/h. It can go 30 km upstream and return downstream to the original point in 4 hours and 30 minutes. Find the speed of the stream.
Sol :Let the speed of the boat be ‘X’ km/ hr, time taken for upstream and downstream be T1 hrs & T2 hrs respectively.
Speed $=\frac{\text { Distance }}{\text { Time }}$
$\mathrm{T} 1=\frac{30}{15+\mathrm{X}}$ ––––––– (i) (Downstream)
$\mathrm{T} 2=\frac{30}{15-\mathrm{X}}$ ––––––– (ii) (Upstream)
Adding equation equation (i) & equation (ii),
$\mathrm{T} 1+\mathrm{T} 2=\frac{30}{15+\mathrm{X}}+\frac{30}{15-\mathrm{X}}$
$\frac{9}{2}=\frac{450-30 X+450+30 X}{225-X^{2}}$
2025 – 9X2 = 1800
9X2 = 225
X2 = 25
X = ±5
∵ speed can’t be negative.
∴ The speed of the stream be 5 Km/hr
Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train $=\frac{132}{x}$ hr
The time taken by the express train $=\left\{\frac{132}{x}+11\right\}$hr
$\mathrm{T} 1+\mathrm{T} 2=\frac{30}{15+\mathrm{X}}+\frac{30}{15-\mathrm{X}}$
$\frac{9}{2}=\frac{450-30 X+450+30 X}{225-X^{2}}$
2025 – 9X2 = 1800
9X2 = 225
X2 = 25
X = ±5
∵ speed can’t be negative.
∴ The speed of the stream be 5 Km/hr
Question 36
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Sol :Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train $=\frac{132}{x}$ hr
The time taken by the express train $=\left\{\frac{132}{x}+11\right\}$hr
Now, according to the question
$\left\{\frac{132}{x+11}\right\}=\frac{132}{x}+1$
After taking L.C.M. of$\frac{132}{x}+1$ and then solving it we get $\frac{132+x}{x}$ .
Now,
$\left\{\frac{132}{x+11}\right\}=\frac{132+x}{x}$
By cross multiplying, we get
132x = x2 + 132x + 11x + 1452
x2 + 11x – 1452 = 0
x2 + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x + 33) (x + 44) = 0
x – 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed of express train is 33 + 11 = 44 km/hr.
Let the present age of Rehman be x years.
So, 3 years age his age was = (x – 3) years
The reciprocal $=\frac{1}{x-3}$
And, after 5 years the age will be = (x + 5) years
The reciprocal $=\frac{1}{x+5}$
$\left\{\frac{132}{x+11}\right\}=\frac{132}{x}+1$
After taking L.C.M. of$\frac{132}{x}+1$ and then solving it we get $\frac{132+x}{x}$ .
Now,
$\left\{\frac{132}{x+11}\right\}=\frac{132+x}{x}$
By cross multiplying, we get
132x = x2 + 132x + 11x + 1452
x2 + 11x – 1452 = 0
x2 + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x + 33) (x + 44) = 0
x – 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed of express train is 33 + 11 = 44 km/hr.
Question 37
The sum of the reciprocals of Rehman's age (in years) 3 years ago and 5 years from now is $\frac{1}{3}$. Find his present age.
Sol :Let the present age of Rehman be x years.
So, 3 years age his age was = (x – 3) years
The reciprocal $=\frac{1}{x-3}$
And, after 5 years the age will be = (x + 5) years
The reciprocal $=\frac{1}{x+5}$
So, according to the question
$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$
$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$
Taking L. C. M. of (x – 3) and (x + 5)
$\frac{(x+5)+x-3}{x^{2}+2 x-15}=\frac{1}{3}$
$\frac{(2 x+2)}{x^{2}+2 x-15}=\frac{1}{3}$
6x + 6 = x2 + 2x – 15
x2 + 2x – 6x – 15 – 6 = 0
x2 – 4x – 21 = 0
x2 – 7x + 3x – 21 = 0
x(x – 7) + 3(x – 7) = 0
(x – 7)(x + 3) = 0
x = 7 and x = –3
x = – 3 is not possible because age cannot be negative.
So, x = 7
Therefore present age of Rehman is 7 years.
Let the father and his son’s age be ‘X’ yrs and ‘Y’ yrs respectively.
X + Y = 35
X × Y = 150
$\mathrm{X}=\frac{150}{\mathrm{Y}}$
$\frac{150}{\mathrm{Y}}+\mathrm{Y}=35$150 + Y2 = 35Y
Y2 – 35Y + 150 = 0
(Y–5) (Y–30)=0
Y=5
the son's age (Y) = 5 yrs and father's age (X) = 30 yrs.
Let the father and his son’s age be ‘X’ yrs and ’(24–X)’ yrs respectively.
As per the question,
$\frac{\{(24-\mathrm{X})(\mathrm{X})\}}{4}-24+\mathrm{X}=9$
24X – X2 – 96 + 4X = 36
X2 – 28X + 132 = 0
X2 – 22X – 6X + 132 = 0
X(X – 22) – 6(X – 22) = 0
(X – 6)(X – 22) = 0
∴X = 22 & boy’age is 2 years.
∴X = 6 & boy’age is 18 years. (practically not possible)
The age of father and his son are 22 years & 2 years respectively.
Let the sister and her sister’s age be ‘X’ yrs and ’(X+5)’ yrs respectively.
As per the question,
X(X + 5) = 104
X2 + 5X – 104 = 0
X2 + 13X – 8X + 132 = 0
X(X + 13) – 8(X + 13) = 0
(X + 13)(X – 8) = 0
∴X = 8 or X = –13
the ages of sisters are 8 years & 13 years respectively, as the age can’t be negative.
Let seven years, age of Swati was x years and age of Varun was 5x2years.
Present age of Swati = (x + 7) years
Present age of Varun = (5x2 + 7) years
Given, after 3 years swati's age will be $\frac{2}{5}$ th of Varun's age.
Age of Swati after 3 years = (x + 7 + 3) years = (x + 10) years
Age of Varun after 3 years = (5x2 + 7 + 3) years = (5x2 + 10) years
Given, age of Swati after 3 years = Two–fifths age of varun after 3 years
$(x+10)=\frac{2}{5} \times\left(5 x^{2}+10\right)$
(x + 10) = 2(x2 + 2)
x + 10 = 2x2 + 4
2x2 – x + 4 – 10 = 0
2x2 – x – 6 = 0
2x2 – 4x + 3x – 6 = 0
2x(x – 2) + 3(x – 2) = 0
x = 2 [ Age can't be negative ]
Therefore, the present age of Swati = (2 + 7) = 9 year and the present age of Varun = 5(2)2+7=27 years.
Let the marks scored in maths be ‘X’.
Marks in English is ‘(40–X)’.
As, per the question,
If he got 3 marks in maths & 4 marks less in English,
Marks in Maths =X+3
Marks in English = 40–X–4 = 36–X
Product = 360
(36 – X)(X + 3) = 360
(36X + 108 – X2 – 3X) = 360
(33X + 108 – X2) = 360
X2 – 33X + 360 – 108 = 0
X2 – 33X + 252 = 0
X2 – 21X – 12X + 252 = 0
X(X – 21) – 12(X – 21) = 0
(X– 12)(X – 21) = 0
X = 12 or 21
∴ If marks in Maths = 12 then marks in English = 40 – 12 = 28
If marks in Maths = 21 then marks in English = 40 – 21 = 19
Let the marks in maths be ‘X’ and English be ‘Y’.
X + Y = 45 equation 1
X = 45 – Y
(X + 1)(Y – 1) = 500
(45 – Y – 1)(Y – 1) = 500
Y2 – 43Y + 456 = 0
By solving this quadratic equation
Y2 – 24Y – 19Y + 456 = 0
Y(Y – 24) – 19(Y – 24) = 0
we get two values of Y
Y1 = 24
Y2 = 19
substitute both this values in equation 1
X1 + 24 = 45
X1 = 45 – 24
= 21
X2 + 19 = 45
X2 = 45 – 19
= 26
the marks in maths is 21, then marks in English is 24 and if the marks in maths is 26, then marks in English is 19.
Let x be the no. of person and y is amount taken by each person.
$\frac{6500}{x}=y$ ……(1)
when 15 more person appear then
$\frac{6500}{x+15}=y-30$ ……(2)
On solving both (1) and (2) equation
$\frac{(x+5)+x-3}{x^{2}+2 x-15}=\frac{1}{3}$
$\frac{(2 x+2)}{x^{2}+2 x-15}=\frac{1}{3}$
6x + 6 = x2 + 2x – 15
x2 + 2x – 6x – 15 – 6 = 0
x2 – 4x – 21 = 0
x2 – 7x + 3x – 21 = 0
x(x – 7) + 3(x – 7) = 0
(x – 7)(x + 3) = 0
x = 7 and x = –3
x = – 3 is not possible because age cannot be negative.
So, x = 7
Therefore present age of Rehman is 7 years.
Question 38
The sum of the ages (in years) of a son and his father is 35 and the their ages is 150. Find their ages.
Sol :Let the father and his son’s age be ‘X’ yrs and ‘Y’ yrs respectively.
X + Y = 35
X × Y = 150
$\mathrm{X}=\frac{150}{\mathrm{Y}}$
$\frac{150}{\mathrm{Y}}+\mathrm{Y}=35$150 + Y2 = 35Y
Y2 – 35Y + 150 = 0
(Y–5) (Y–30)=0
Y=5
the son's age (Y) = 5 yrs and father's age (X) = 30 yrs.Question 39
If a boy's age and his father's age amount together to 24 years. Fourth pan i product of their ages exceeds the boy's age by 9 years. Find how old they are?
Sol :Let the father and his son’s age be ‘X’ yrs and ’(24–X)’ yrs respectively.
As per the question,
$\frac{\{(24-\mathrm{X})(\mathrm{X})\}}{4}-24+\mathrm{X}=9$
24X – X2 – 96 + 4X = 36
X2 – 28X + 132 = 0
X2 – 22X – 6X + 132 = 0
X(X – 22) – 6(X – 22) = 0
(X – 6)(X – 22) = 0
∴X = 22 & boy’age is 2 years.
∴X = 6 & boy’age is 18 years. (practically not possible)
The age of father and his son are 22 years & 2 years respectively.Question 40
The product of the ages of two sisters is 104. The difference between their ages is 5. Find their ages.
Sol :Let the sister and her sister’s age be ‘X’ yrs and ’(X+5)’ yrs respectively.
As per the question,
X(X + 5) = 104
X2 + 5X – 104 = 0
X2 + 13X – 8X + 132 = 0
X(X + 13) – 8(X + 13) = 0
(X + 13)(X – 8) = 0
∴X = 8 or X = –13
the ages of sisters are 8 years & 13 years respectively, as the age can’t be negative.Question 41
Seven years ago Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two–fifth of Varun's age. Find their present ages.
Sol :Let seven years, age of Swati was x years and age of Varun was 5x2years.
Present age of Swati = (x + 7) years
Present age of Varun = (5x2 + 7) years
Given, after 3 years swati's age will be $\frac{2}{5}$ th of Varun's age.
Age of Swati after 3 years = (x + 7 + 3) years = (x + 10) years
Age of Varun after 3 years = (5x2 + 7 + 3) years = (5x2 + 10) years
Given, age of Swati after 3 years = Two–fifths age of varun after 3 years
$(x+10)=\frac{2}{5} \times\left(5 x^{2}+10\right)$
(x + 10) = 2(x2 + 2)
x + 10 = 2x2 + 4
2x2 – x + 4 – 10 = 0
2x2 – x – 6 = 0
2x2 – 4x + 3x – 6 = 0
2x(x – 2) + 3(x – 2) = 0
x = 2 [ Age can't be negative ]
Therefore, the present age of Swati = (2 + 7) = 9 year and the present age of Varun = 5(2)2+7=27 years.
Question 42 A
In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of his marks would have been 360. Find his marks in two subjects separately.
Sol :Let the marks scored in maths be ‘X’.
Marks in English is ‘(40–X)’.
As, per the question,
If he got 3 marks in maths & 4 marks less in English,
Marks in Maths =X+3
Marks in English = 40–X–4 = 36–X
Product = 360
(36 – X)(X + 3) = 360
(36X + 108 – X2 – 3X) = 360
(33X + 108 – X2) = 360
X2 – 33X + 360 – 108 = 0
X2 – 33X + 252 = 0
X2 – 21X – 12X + 252 = 0
X(X – 21) – 12(X – 21) = 0
(X– 12)(X – 21) = 0
X = 12 or 21
∴ If marks in Maths = 12 then marks in English = 40 – 12 = 28
If marks in Maths = 21 then marks in English = 40 – 21 = 19
Question 42 B
In a class test, the sum of Gagan marks in Mathematics and English is 45. If he had 1 more mark in Mathematics and 1 less in English, the product of marks would have been 500. Find the original marks obtained by Gagan in Mathematics and English separately.
Sol :Let the marks in maths be ‘X’ and English be ‘Y’.
X + Y = 45 equation 1
X = 45 – Y
(X + 1)(Y – 1) = 500
(45 – Y – 1)(Y – 1) = 500
Y2 – 43Y + 456 = 0
By solving this quadratic equation
Y2 – 24Y – 19Y + 456 = 0
Y(Y – 24) – 19(Y – 24) = 0
we get two values of Y
Y1 = 24
Y2 = 19
substitute both this values in equation 1
X1 + 24 = 45
X1 = 45 – 24
= 21
X2 + 19 = 45
X2 = 45 – 19
= 26
the marks in maths is 21, then marks in English is 24 and if the marks in maths is 26, then marks in English is 19.Question 43
Rs. 6500 were divided equally among a certain number of persons. Had there bees 1 15 more persons, each would have got Rs. 30 less. Find the original number of persons.
Sol :Let x be the no. of person and y is amount taken by each person.
$\frac{6500}{x}=y$ ……(1)
when 15 more person appear then
$\frac{6500}{x+15}=y-30$ ……(2)
On solving both (1) and (2) equation
subtracting (2) from (1)
$\frac{6500}{x}-\frac{6500}{x+15}=30$
$\left(\frac{1}{x}-\frac{1}{x+15}\right) 6500=30$
$\frac{x+15-x}{x^{2}+15 x}=\frac{30}{6500}$
3250 = x2 + 15x
x2 + 15x – 3250 = 0
$\frac{6500}{x}-\frac{6500}{x+15}=30$
$\left(\frac{1}{x}-\frac{1}{x+15}\right) 6500=30$
$\frac{x+15-x}{x^{2}+15 x}=\frac{30}{6500}$
3250 = x2 + 15x
x2 + 15x – 3250 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(15) \pm \sqrt{(15)^{2}-4(1)(-3250)}}{2(1)}$
$\frac{-15 \pm \sqrt{225+13000}}{2}$
$\frac{-15 \pm \sqrt{13225}}{2}$
$\frac{-15 \pm 115}{2}$
X = – 65 or X = 50
The persons are 50 (As positive values are only considered).
Let x be the no. of students and y is the number of apples taken by each person.
$\frac{300}{x}=y$ ––––––(1)
when 10 more students appear then,
$\frac{300}{x+10}=y-1$ –––––(2)
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(15) \pm \sqrt{(15)^{2}-4(1)(-3250)}}{2(1)}$
$\frac{-15 \pm \sqrt{225+13000}}{2}$
$\frac{-15 \pm \sqrt{13225}}{2}$
$\frac{-15 \pm 115}{2}$
X = – 65 or X = 50
The persons are 50 (As positive values are only considered).Question 44
300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.
Sol :Let x be the no. of students and y is the number of apples taken by each person.
$\frac{300}{x}=y$ ––––––(1)
when 10 more students appear then,
$\frac{300}{x+10}=y-1$ –––––(2)
On solving both (1) and (2) equation
subtracting (2) from (1)
$\frac{300}{x}-\frac{300}{x+10}=1$
$\left(\frac{1}{x}-\frac{1}{x+10}\right) 300=1$
$\frac{x+10-x}{x^{2}+10 x}=\frac{1}{300}$
3000 = x2 + 10x
x2 + 10x – 3000 = 0
$\frac{300}{x}-\frac{300}{x+10}=1$
$\left(\frac{1}{x}-\frac{1}{x+10}\right) 300=1$
$\frac{x+10-x}{x^{2}+10 x}=\frac{1}{300}$
3000 = x2 + 10x
x2 + 10x – 3000 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(10) \pm \sqrt{(10)^{2}-4(1)(-3000)}}{2(1)}$
$\frac{-10 \pm \sqrt{100+12000}}{2}$
$\frac{-10 \pm \sqrt{12100}}{2}$
$\frac{-10 \pm 110}{2}$
X = – 60 or X = 50
The number of students are 50. (as only positive values are considered).
Let x be the no. of books and y is the cost of each book.
$\frac{1200}{x}=y$ ––––––(1)
when 10 more students appear then,
$\frac{1200}{x+10}=y-20$ ––––(2)
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(10) \pm \sqrt{(10)^{2}-4(1)(-3000)}}{2(1)}$
$\frac{-10 \pm \sqrt{100+12000}}{2}$
$\frac{-10 \pm \sqrt{12100}}{2}$
$\frac{-10 \pm 110}{2}$
X = – 60 or X = 50
The number of students are 50. (as only positive values are considered).Question 45
A shopkeeper buys a number of books for Rs. 1200. If he had bought 10 more books for the same amount, each book would have cost Rs. 20 less. How many books did he buy?
Sol :Let x be the no. of books and y is the cost of each book.
$\frac{1200}{x}=y$ ––––––(1)
when 10 more students appear then,
$\frac{1200}{x+10}=y-20$ ––––(2)
On solving both (1) and (2) equation
subtracting (2) from (1)
$\frac{1200}{x}-\frac{1200}{x+10}=20$
$\left(\frac{1}{x}-\frac{1}{x+10}\right) 1200=20$
$\frac{x+10-x}{x^{2}+10 x}=\frac{20}{1200}$
600 = x2 + 10x
x2 + 10x – 600 = 0
subtracting (2) from (1)
$\frac{1200}{x}-\frac{1200}{x+10}=20$
$\left(\frac{1}{x}-\frac{1}{x+10}\right) 1200=20$
$\frac{x+10-x}{x^{2}+10 x}=\frac{20}{1200}$
600 = x2 + 10x
x2 + 10x – 600 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(10) \pm \sqrt{(10)^{2}-4(1)(-600)}}{2(1)}$
$\frac{-10 \pm \sqrt{100+2400}}{2}$
$\frac{-10 \pm \sqrt{2500}}{2}$
$\frac{-10 \pm 50}{2}$
X = – 30 or X = 20
∴ The number of books, he purchased are 20. (as only positive values are considered).
Let the total number of camels be x.
Number of camels in forest $=\frac{x}{4}$ .
Number of camels gone to mountains = 2√x
Remaining camels on bank of the river = 15
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(10) \pm \sqrt{(10)^{2}-4(1)(-600)}}{2(1)}$
$\frac{-10 \pm \sqrt{100+2400}}{2}$
$\frac{-10 \pm \sqrt{2500}}{2}$
$\frac{-10 \pm 50}{2}$
X = – 30 or X = 20
∴ The number of books, he purchased are 20. (as only positive values are considered).
Question 46
One–fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of the river. Find the total number of camels.
Sol :Let the total number of camels be x.
Number of camels in forest $=\frac{x}{4}$ .
Number of camels gone to mountains = 2√x
Remaining camels on bank of the river = 15
Total camels $=\frac{x}{4}+2 \sqrt{x}+15$
$x=\frac{x}{4}+2 \sqrt{x}+15$
4x = x + 8√x + 60
3x – 60 = 8√x
$x=\frac{x}{4}+2 \sqrt{x}+15$
4x = x + 8√x + 60
3x – 60 = 8√x
Squaring both the sides,
9x2 + 3600 – 360x = 64x
9x2 – 424x + 3600 = 0
On simplifying further,
9x2 – 100x – 324x + 3600 = 0
x(9x – 100) – 36(9x – 100) = 0
(x – 36)(9x – 100) = 0
∴ x = $\frac{100}{9}$or X = 36
∴ The number of camels is 36. (As whole values are only considered)
Amount for the booking of hotel = Rs. 1200
Let there be x tourists
When all the tourist paid money, each share = Rs. $\frac{1200}{x}$
When 3 members failed to pat, its Rs. $\frac{1200}{x-3}$
9x2 + 3600 – 360x = 64x
9x2 – 424x + 3600 = 0
On simplifying further,
9x2 – 100x – 324x + 3600 = 0
x(9x – 100) – 36(9x – 100) = 0
(x – 36)(9x – 100) = 0
∴ x = $\frac{100}{9}$or X = 36
∴ The number of camels is 36. (As whole values are only considered)
Question 47
A party of tourists booked a room in a hotel for Rs. 1200. Three of the members failed to pay. As a result, others had to pay Rs. 20 more (each). How many tourists were there in the party?
Sol :Amount for the booking of hotel = Rs. 1200
Let there be x tourists
When all the tourist paid money, each share = Rs. $\frac{1200}{x}$
When 3 members failed to pat, its Rs. $\frac{1200}{x-3}$
So, according to the question:
$\frac{1200}{x-3}-\frac{1200}{x}=20$
$\frac{1200 x-1200 x+3600}{x(x-3)}=20$
3600 = 20 × x(x – 3)
3600 = 20x2– 60x
20x2 – 60x – 3600 = 0
$\frac{1200}{x-3}-\frac{1200}{x}=20$
$\frac{1200 x-1200 x+3600}{x(x-3)}=20$
3600 = 20 × x(x – 3)
3600 = 20x2– 60x
20x2 – 60x – 3600 = 0
On simplifying further,
x2 – 3x – 180 = 0
x2 – 15x + 12x – 180 =0
x(x – 15) + 12(x – 15) = 0
Therefore, x = –12 or x = 15
∴ The number of camels is 15. (As positive values are only considered)
Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+5)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{X}+\frac{1}{X+5}=\frac{1}{6}$
$\frac{X+5+X}{X^{2}+5 X}=\frac{1}{6}$
$\frac{(2 X+5)}{X^{2}+5 X}=\frac{1}{6}$
12X + 30 = X2 + 5X
X2 – 7X – 30 = 0
x2 – 3x – 180 = 0
x2 – 15x + 12x – 180 =0
x(x – 15) + 12(x – 15) = 0
Therefore, x = –12 or x = 15
∴ The number of camels is 15. (As positive values are only considered)
Question 48
Two pipes running together can fill a cistern in 6 minutes. If one pipe takes 5 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Sol :Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+5)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{X}+\frac{1}{X+5}=\frac{1}{6}$
$\frac{X+5+X}{X^{2}+5 X}=\frac{1}{6}$
$\frac{(2 X+5)}{X^{2}+5 X}=\frac{1}{6}$
12X + 30 = X2 + 5X
X2 – 7X – 30 = 0
On factorising the same.
X2 – 10X + 3X – 30 = 0
X(X – 10) + 3(X – 10) = 0
(X – 10)(X + 3) = 0
∴ X = – 3 or X = 10.
Then the first pipe will fill the cistern in ‘X’ minutes i.e., 10 minutes and the second pipe will fill the cistern in ‘(X+5)’ minutes i.e., 15 minutes.
Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+1)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{\mathrm{X}}+\frac{1}{\mathrm{X}+1}=\frac{11}{30}$
$\frac{X+1+X}{X^{2}+x}=\frac{11}{30}$
$\frac{(2 \mathrm{X}+1)}{\mathrm{X}^{2}+\mathrm{X}}=\frac{11}{30}$
60X + 30 = 11X2 + 11X
11X2 – 49X – 30 = 0
X2 – 10X + 3X – 30 = 0
X(X – 10) + 3(X – 10) = 0
(X – 10)(X + 3) = 0
∴ X = – 3 or X = 10.
Then the first pipe will fill the cistern in ‘X’ minutes i.e., 10 minutes and the second pipe will fill the cistern in ‘(X+5)’ minutes i.e., 15 minutes.
Question 49
Two pipes running together can fill a cistern in $\frac{30}{11}$ minute. If one pipe takes 1 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Sol :Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+1)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{\mathrm{X}}+\frac{1}{\mathrm{X}+1}=\frac{11}{30}$
$\frac{X+1+X}{X^{2}+x}=\frac{11}{30}$
$\frac{(2 \mathrm{X}+1)}{\mathrm{X}^{2}+\mathrm{X}}=\frac{11}{30}$
60X + 30 = 11X2 + 11X
11X2 – 49X – 30 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-49) \pm \sqrt{(-49)^{2}-4(11)(-30)}}{2(11)}$
$\frac{49 \pm \sqrt{2401+1320}}{22}$
$\frac{49 \pm \sqrt{3721}}{22}$
$\frac{49 \pm 61}{22}$
X = $-\frac{6}{11}$ or X = 5
∴ The time required to fill the cistern is 5 & 6 minutes respectively. (as only positive values are considered).
Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+3)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{X}+\frac{1}{X+3}=\frac{13}{40}$
$\frac{X+3+X}{X^{2}+3 X}=\frac{13}{40}$
$\frac{(2 \mathrm{X}+3)}{\mathrm{X}^{2}+3 \mathrm{X}}=\frac{13}{40}$
80X + 120 = 13X2 + 39X
13X2 – 41X – 120 = 0
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-49) \pm \sqrt{(-49)^{2}-4(11)(-30)}}{2(11)}$
$\frac{49 \pm \sqrt{2401+1320}}{22}$
$\frac{49 \pm \sqrt{3721}}{22}$
$\frac{49 \pm 61}{22}$
X = $-\frac{6}{11}$ or X = 5
∴ The time required to fill the cistern is 5 & 6 minutes respectively. (as only positive values are considered).
Question 50
Two pipes running together can fill a cistern in $\frac{40}{13}$minutes. If one pipe takes 3 minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
Sol :Let the first pipe fill the cistern in ‘X’ minutes, then the second pipe requires ‘(X+3)’ minutes to fill it.
Applying the concept of Unitary Method,
In one minute, both pipes will fill the part of cistern as below:
$\frac{1}{X}+\frac{1}{X+3}=\frac{13}{40}$
$\frac{X+3+X}{X^{2}+3 X}=\frac{13}{40}$
$\frac{(2 \mathrm{X}+3)}{\mathrm{X}^{2}+3 \mathrm{X}}=\frac{13}{40}$
80X + 120 = 13X2 + 39X
13X2 – 41X – 120 = 0
On applying Sreedhracharya formula
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-41) \pm \sqrt{(-41)^{2}-4(13)(-120)}}{2(13)}$
$\frac{41 \pm \sqrt{1681+6240}}{26}$
$\frac{41 \pm \sqrt{7921}}{26}$
$\frac{41 \pm 89}{26}$
X = $-\frac{24}{13}$ or X = 5
∴ The time required to fill the cistern is 5 & 8 minutes respectively. (as only positive values are considered).
$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\frac{-(-41) \pm \sqrt{(-41)^{2}-4(13)(-120)}}{2(13)}$
$\frac{41 \pm \sqrt{1681+6240}}{26}$
$\frac{41 \pm \sqrt{7921}}{26}$
$\frac{41 \pm 89}{26}$
X = $-\frac{24}{13}$ or X = 5
∴ The time required to fill the cistern is 5 & 8 minutes respectively. (as only positive values are considered).
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
No comments:
Post a Comment