CAT-Numerical Ability-3

CAT – Numerical Ability – 3

1. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m^3) is:
a) 4830                  b) 5120                 c)6420                   d)8960

Ans: (b)

The side of the rectangular sheet is 48 m and 36 m. A square of 8 m is cut from all the corners of this sheet. Thus 8 m + 8 m = 16 m is reduced from all the side. The new side will be of dimension 32 m and 20 m. This is folded in to an open box the height of which is 8 m.

Hence the volume of the box is -> 32 x 20 x 8 = 5120 m^3

2. A group of students decided to collect as many paise from each member of group equalling to the number of members in the group.  If the total collection amounts to Rs. 59.29, then the number of members in the group is:
a) 57                      b) 67                      c) 77                       d) 87
Ans: (c)
The total amount collected is Rs 59.29 or 5929 paise. Since the number of members in the group and the number of paise are equal, the number of members is equal to √5929 -> 77.

3. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
a) 9                         b) 10                      c) 12                       d) 20
Ans: (b)

In one hour the bus runs 9 km less due to stoppages. The time taken for these stoppages is

(9/54) x 60 -> 10 minutes.  

4. A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
a) 14 km               b) 15 km               c)16 km                                d) 17km
Ans: (c)
Let the distance travelled by foot be ‘X’. Then the distance travelled by cycle is (61 – X). Now we have an equation X/4 + (61-X)/ 9 = 9 solving we get the value of ‘X’ as 16 km.

5. The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?
a) 69                      b) 78                      c) 96                       d) Cannot be determined
Ans: (d)

The question would appear easy and the natural tendency would be to mark (c) as the answer. But, the two digit number could be either 96 or 69. Hence the answer is cannot be determined.

6. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
a) Rs. 4991.         B) Rs. 5991          c) Rs. 6001           d) Rs. 6991
Ans: (a)

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
For average sale of Rs. 6500 for six months the total sales for six months should be 6500 x 6 = Rs 39000. Hence the sixth month sale should be 39000 – 34009 = Rs. 4991.

7. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is :

a) 74                      b) 94                      c) 184                    d) 364
Ans: (d)
The L.C.M. of 6, 9, 15, and 18 is 90.

The least number which is a multiple of 7 leaving a reminder of 4 be (90x + 4) By trial we can arrive the least value for ‘x’ as 4. Hence the least number is (90x4) + 4 = 364.  

8. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
a) 2010                  b) 2011                 c) 2012                  d) 2013
Ans: (b)

Let us assume that commodity X will be 0.40 paise more than commodity Y after ‘Z’ years.

Then we have an equation -> (4.20 + 0.40Z) – (6.30 + 0.15Z) = 0.40 paise.

Solving we get the value of ‘Z’ as 10 years. Hence, 10 years after 2001 ie 2011 the price of commodity ‘X’ will be 0.40 paise more than the price f the commodity ‘Y’.

  
9. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
a) 4                         b) 5                        c) 6                         d) 8
Ans: (a)

Let ‘N’ be the HCF of 1305, 4665, and 6905. Then,

HCF of (4665 – 1305), (6905 – 4665), and (6905 – 1305) -> (3360), (2240) and (5600) is 1120

Thus N = 1120 and the sum of the digits 1 + 1 + 2 + 0 = 4
 
10. Sum of five consecutive even numbers is 380. What is the second number in ascending order?
a) 76                      b) 78                      c) 74                       d) 72
Ans: (c)
The sum of the five consecutive even numbers is 360. The average is 360/5 = 76 which is the middle number and three in the ascending order. Hence, the second number is -> 76 – 2 = 74.