Fractions
 A man reads 3/8 of a book on a day and 4/5 of the remainder on the second day. If the number of pages still unread are 40, then how many pages did the book contain?

 300
 500
 320
 350

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Let the number of pages be p.
Number of pages read on first day = 3p/8
Remaining pages = p  3p/8 = (8p  3p)/8 = 5p/8
Number of pages read on second day = 4/5 x 5p/8 = p/2
Now , remaining pages = p  [ 3p/8 + p/2]Correct Option: C
Let the number of pages be p.
Number of pages read on first day = 3p/8
Remaining pages = p  3p/8 = (8p  3p)/8 = 5p/8
Number of pages read on second day = 4/5 x 5p/8 = p/2
Now , remaining pages = p  [ 3p/8 + p/2]
= p  [(3p + 4p)/8] = [p 7p/ 8]
= [(8p  7p)/ 8] = p/8
According to the question
p/8 = 40
⇒ p = 40 x 8
∴ p = 320
 If the numerator of a fraction is increased by 200% and the denominator of the fraction is increased by 150%, the resultant fraction is 9/35. What is the original fraction ?

 3/10
 2/15
 3/16
 2/7
 None of the above

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Let the original fraction be p/q.
Numerator is increased by 200%.
∴ Numerator = p + 200% of p
= p + 200p/100
= p + 2p
= 3p
and denominator of the fraction is increased by 150%.
Denominator = q + 150q/100Correct Option: E
Let the original fraction be p/q.
Numerator is increased by 200%.
∴ Numerator = p + 200% of p
= p + 200p/100
= p + 2p
= 3p
and denominator of the fraction is increased by 150%.
Denominator = q + 150q/100
= (100q + 150q)/100
= 250q/100
= 5q/2
Then, according to the question,
3p/5q/2 = 9/35
3p x 2/5q = 9/35
6p/5q = 9/35
∴ p/q = 9 x 5/35 x 6
p/q = 3/14
 A, B, C and D purchase a gift worth ₹ 60. A pays 1/2 of what others are paying, B pays 1/3rd of what other are paying and C pays 1/4th of what others are paying. What is the amount paid by D ?

 14
 15
 16
 13

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Let A, B, C and D pay ₹x, ₹y, ₹z and ₹a
According to the question,
x = 1/2(y + z + a)
2x = (y + z + a) ..........(i)
y = 1/3(x + z + a)
3y = (x + z + a) ...........(ii)
z = 1/4(x + y + a)
4z = (x + y + a) ..........(iii)
Also, x + y + z + a = 60 ...............(iv)Correct Option: D
Let A, B, C and D pay ₹x, ₹y, ₹z and ₹a
According to the question,
x = 1/2(y + z + a)
2x = (y + z + a) ..........(i)
y = 1/3(x + z + a)
3y = (x + z + a) ...........(ii)
z = 1/4(x + y + a)
4z = (x + y + a) ..........(iii)
Also, x + y + z + a = 60 ...............(iv)
Now, put the value of x + y + a = 4z from (iii) in (iv)
Then, 4z + z = 60 ⇒ 5z = 60
∴ z = 12
Similarly, on putting the value of x + z + a = 3y from (ii) in (iv), we get
3y + y = 60 = ⇒ 4y = 60
∴ y = 15
Again, on putting the value of (y + z + a) = 2x from (i) in (iv), we get
2x + x = 60 ⇒ 3x = 60
&ther4; x = 20
Now, x + y + z + a = 60
On putting the value of x, y and z, we get
12 + 15 + 20 + a = 60
∴ a = 60  47 = ₹ 13
 The Numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator, then find the fraction.

 3/7
 4/8
 2/7
 3/8

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Let denominator of fraction = x
Then, numerator = x  4
∴ Fraction = x  4/x
Now, according to the question,
8[(x  4)  2] = (x + 1)Correct Option: A
Let denominator of fraction = x
Then, numerator = x  4
∴ Fraction = x  4/x
Now, according to the question,
8[(x  4)  2] = (x + 1)
(x 4)  2= (x + 1)/8
⇒ x  6 = x + 1/8
⇒ 8x  48 = x + 1
⇒ 8x  x = 48 + 1
⇒ 7x = 49
⇒ x = 49/7
∴ x = 7
∴ Fraction = (x  4)/x
Put the value of x.
∴ Fraction = (7  4)/7 =3/7
 783 ÷ 9 ÷ 0.75 = ?

 130
 124
 118
 116
 None of the above

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? = 783/9 x 0.75 = 116
Correct Option: D
? = 783/9 x 0.75 = 116