# Maths Quiz For IBPS PO: Part 6

### Question 1:

A sum of money was distributed equally among a certain number of children . If there were 5children less , each would have received a rupee more .But if there were 15children more , each would have received Rs.2 less .Find the sum of money distributed.
a)120
b)125
c)100
d) 360

### Question 2:

In a class test, A scored 51 marks. The test had 4 sections. In the second section, A scored 7 marks more than in the first section, and in the third section A Scored 7 marks more than in the second section. If A did not score at all in the fourth section , how much did A score in the second section ?
a) 15
b) 10
c) 24
d) 17

### Question 3:

A three digit number is such that , when divided by 13 , it yields twice the digit in the unit’s place as quotient and zero as remainder .We also know that the digit in the hundreds’place is not a prime number and that the digit in the tens’place is zero .Find the digit in the untis’place.
a) 3
b)4
c)1
d)9

## DATA SUFFICIENCY :

Directions: Each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and give answer-
a) If the data in both statements I and II together are not sufficient to answer the question.
b) If the data in both the statement I and II together are necessary to answer the question.
c) If either of the data in statement I alone or in statement II alone are sufficient to answer the question.
d) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question
e) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.

### Question 4:

Is the product of two given consecutive natural numbers greater than 92 ?
(i) The difference between the square of the larger number and the larger number is equal to 90.
(ii) The smaller of the two numbers , when added be a squared even integer and then divided by 2 , gives the result 6.5

### Question 5:

Certain quantities of two mixtures X and Y are mixed to get a new mixture containing 20% alcohol. Find the percentage of alcohol. Find the percentage of alcohol in mixture X .
(i) Mixture Y contains twice as much alcohol as mixture X
(II) Qunatity of mixture X is twice that of mixture Y

## Explanation

#### 1) d

Let the amount each child has be Rs.y and the number of children be x
(x-5)(y+1) = xy ….. (i)
(x+15) (y-2) = xy (ii)
From (i) xy+x-5y-5 = xy
X – 5y = 5 …… (iii)
From (ii) xy -2x +15y – 30 = xy
-2x +15y = 30 …… (iv)
Multiplying (iii) by 2
2x-10y = 10 ….. (v)
Adding (iv) and (v), we get 5y = 40
so y =8
substituting the value of ‘y’in equation(iii)
x-40 = 5
x=45
amount distributed = xy = 45* 8 = 360
Hence , (d)360

#### 2) d

If A Scored x marks in section 2 , he scored (x-7) marks I nsection 1 and (x+7) marks in section 3
Therefore , (x-7) + x + (x+7) + 0 =51
3x=51
X=17

#### 3) b

Let the digit in the units place be x .and that in the hundreds’place be y.
So the number be 100y + x
(100y + x ) = 13*2x +0
i.e. 100y+x = 26X so x=4y
As x and y are single digits , y can only take the values of 1 and 2 . but as y is not prime y=1 and hence x=4 , Hence (b) 4

#### 4) e

Let the smaller and larger numbers be n and n+1 respectively …
Statement (I ) : (n+1)² – (n-1) = 90
Or n² +2n +1 –n-1 =90
Or n² +n = 90
Or n(n+1) = 90
Or (smaller number ) * ( larger number ) = 90
So , I alone is not sufficient .
Statement (II) : Let n bethe smaller of the two numbers then
N+2a² / 2 = 6.5
N+(2a)² = 13
The only squared even integer less than 13 is 4
N=9
Hence , II alone is sufficient .

#### 5)

Any of the statements is individually insufficient to answer the question ..
Combining statements I and ii
Let y and 2y be the quantities of mixtures Y and X respectively , and X and 2x be the quantities of alcohol in mixture X and Y , respectively .
By allegation :
x/2y * 100 2x/y *100
20
(100x/2y – 20 ) / (20-200x/y) = ½
Thus ,we can find

x/2y *100 i.e. the required value ,Hence (b) both statements needed .

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