MAT (Stage-2) Mock Test

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MAT (Stage-2) Mock Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following options should replace the question mark in the final figure of the series given below?

A
14

B
40

C
58

D
6

Solution
Question 2
1 / -0

Which of the following options should replace the question mark in the final figure of the series given below?

A
48

B
14

C
2

D
98

Solution
Question 3
1 / -0

The difference between the squares of two consecutive odd integers is always divisible by

A
3

B
6

C
7

D
8

Solution

Let the two consecutive odd integers be (2x + 3) and (2x + 1).
Then,
(2x + 3)²- (2x + 1)²
= 4x² + 9 + 12x - 4x² - 1 - 4x
= 8 + 8x
= 8(1 + x), which is divisible by 8
Question 4
1 / -0

A number, when divided by 114, leaves a remainder of 21. If the same number is divided by 19, the remainder will be

A
1

B
21

C
7

D
2

Solution

Any number divisible by 114 will also be divisible by 19 because 114 is a multiple of 19.
When the number is divided by 114, it leaves a remainder of 21.
If 21 is divided by 19, then the remainder will be 2.
Question 5
1 / -0

Two out of six papers for an examination are of physics. What are the number of ways in which the papers can be arranged so that two physics papers are not together?

A
348

B
480

C
740

D
800

Solution

Arrangement of total number of papers = 6! or 720
Number of ways when two physics papers are set together = 5! × 2 = 240
So, required number of ways = 720 - 240 = 480
Question 6
1 / -0

The average monthly rainfall for a year in Tokyo is 2.7 inches. The average for the first 7 months is 1.1 inches less than the annual average. If the total rainfall for the next 4 months is 20.8 inches, calculate the rainfall in the last month.

A
0.4 inches

B
0.8 inches

C
1.4 inches

D
2.7 inches

Solution

Annual rainfall = 2.7 × 12 = 32.4 inches
Total rainfall of first 7 months = (2.7 - 1.1) × 7 = 11.2 inches
Rainfall in last month = annual rainfall – [total rainfall of first 7 and next 4 months]
= 32.4 – [11.2 + 20.8] = 32.4 – 32 = 0.4 inches
Question 7
1 / -0

The average score of 60 students in a class is 32.5. What will be the new average if the score of one student is altered from 88 to 78 and the score of another is altered from 13 to 43?

A
32.17

B
33.96

C
32.83

D
33.17

Solution

Total score of 60 students = 60 × 32.5 = 1950
Decrease in the score of first student = 88 - 78 = 10
Increase in the score of second student = 43 - 13 = 30
∴ Newtotal score = 1950 - 10 + 30 = 1970
New average = 1970/60
= 32.83
Question 8
1 / -0

In 21 kg of an alloy, the ratio of copper to zinc by mass is 5 : 2. If 4 kg of copper is added to the alloy, then what will be the new ratio of zinc to copper?

A
16 : 9

B
4 : 3

C
6 : 19

D
3 : 4

Solution

Mass of copper present in 21 kg of alloy = [5/(5 + 2)] × 21 = 15 kg
Mass of zinc = 21 - 15 = 6 kg
New mass of copper = 15 + 4 = 19 kg
Hence, ratio of zinc to copper by mass = 6 : 19
Question 9
1 / -0

Rs. 1400 is divided among three labourers: K, L and M in such a way that the ratio of L's share and K's share is 3 : 2 and the ratio of M's share and L's share is 5 : 4.

Which of the following is K's share?

A
Rs. 220

B
Rs. 320

C
Rs. 120

D
Rs. 160

Solution

Let the shares of K, L and M be x, y and z, respectively.
Given: x : y = 2 : 3 = 8 : 12
y : z = 4 : 5 = 12 : 15
Hence, x : y : z = 8 : 12 : 15
Therefore, K's share = [8/(8 + 12 + 15)] × 1400 = 8/35 × 1400 = Rs. 320
Question 10
1 / -0

Suraj spends 30% of his monthly income on food articles, 40% of remaining on clothes, and saves 50% of the remaining income. If his monthly salary is Rs. 18,400, how much money does he save every month?

A
Rs. 3864

B
Rs. 4264

C
Rs. 3624

D
Rs. 7728

Solution

Income spent on food articles = 30/100 × 18,400 = 5520
Remaining income = Rs. 12,880
Income spent on clothes = 40/100 × 12,800 = Rs. 5152
Remaining income = Rs. 7728
Savings = 50/100 × 7728 = Rs. 3864
Question 11
1 / -0

The monthly income of a person was Rs. 1350 and his monthly expenses were Rs. 900. Next year, his income increased by 14% and expenses decreased by 7% only. Find the percentage increase in the savings.

A
7%

B
21%

C
28%

D
56%

Solution

Old savings = Rs. (1350 - 900) = Rs. 450
New income = 1350 + 14/100 × 1350 = 1350 + 189 = Rs. 1539
New expenses = 900 - 7/100 × 900= Rs. 837
New savings = Rs. (1539 - 837) = Rs. 702
%age of increased savings = 702 - 450/450 × 100 = 252/450 × 100= 56%
Question 12
1 / -0

A can do a piece of work in 20 days, while B can do it in 12 days. B works for 9 days and then leaves. In how many days can A alone finish the remaining work?

A
3 days

B
5 days

C
7 days

D
11 days

Solution

B's 1 day's work = 1/12
∴ B's 9 days' work = 9/12
Remaining work = 1 - 9/12 = 3/12
Now, A takes 20 days to complete the whole work.
Time taken by A to complete 3/12 of the work = 20 × 3/12 = 5 days
Question 13
1 / -0

Mahesh and Umesh can complete a piece of work in 10 days and 15 days, respectively. Umesh starts the work and after 5 days, Mahesh also joins him. In how many days will the work be completed?

A
7 days

B
9 days

C
11 days

D
None of these

Solution
Question 14
1 / -0

The speed of a boat in still water is 10 m/s and the speed of stream is 6 m/s. If the boat goes upstream and comes back downstream, then what is the ratio of the time taken to cover a particular stretch of distance in each direction?

A
4 : 1

B
1 : 4

C
2 : 1

D
None of these

Solution

Let the distance be x m.
Speed of boat upstream = 10 - 6 = 4 m/s
Speed of boat downstream = 10 + 6 = 16 m/s
Time taken by boat downstream = x/16 s
Time taken by boat upstream = x/4 s
Required ratio = x/4 : x/16
= 4 : 1
Question 15
1 / -0

Anil and Sunil start from the same point to run in opposite directions around a circular race course, which is 550 m in circumference. Anil does not start till Sunil runs 100 m. They pass each other when Anil has run a distance of 250 m. What will be the distance between them when the first one reaches the starting point?

A
10 m

B
25 m

C
35 m

D
None of these

Solution

The diagrammatic representation of the situation is as shown.

Let S be the starting point.Let F be the point where they first meet.
Anil has covered a distance of 250 m Sunil has covered a distance of 300 m.

Now, as Sunil had a 100 m lead, in the time Anil covers 250 m, Sunil covers 200 m.
Now Anil has to cover distance of 300 m to reach the starting point.

Sunil has to cover a distance of 250 m.
By the time Anil covers a distance of 300 m, Sunil covers a distance of (300/250 × 250) m = 240 m.

Thus, Anil will reach the starting point first.
As Sunil had to cover 250 m to reach the starting point, he would lag by 10 m to reach the starting point by the time Anil would have reached the finishing point.

Thus, option (1) is correct.