Problem 1 :
Find the value of (100 + 250 ÷ 10) × 3.
(1) 105 (3) 325 
(2) 175 (4) 375 
Solution :
= (100 + 25) × 3
= 125 × 3
= 375
So, option (4) is correct.
Problem 2 :
Which of the following numbers when rounded to the nearest hundred becomes 61,400 ?
(1) 61,349 (3) 61,450 
(2) 61,449 (4) 61,495 
Solution :
First consider option (1).
61,349  Nearest Hundreds : 61,300
Consider option (2).
61,449  Nearest Hundreds : 61,400
Consider option (3).
61,450  Nearest Hundreds : 61,500
Consider option (4).
61,495  Nearest Hundreds : 61,500
So, option (2) is correct.
Problem 3 :
How many sixths are there in 4 5/6 ?
(1) 29 (3) 15 
(2) 26 (4) 5 
Solution :
So, 29 sixths are there in 4 5/6.
So, option (1) is correct.
Problem 4 :
Which of the following is not equal to 1/2 ?


Solution :
Consider option (1). Consider option (3). 
Consider option (2). Consider option (4). 
Hence, option (2) is correct.
Problem 5 :
What is the value of X in the scale below ?
(1) 5.04 (3) 5.2 
(2) 5.02 (4) 5.4 
Solution :
Between 4.8 and 4.9, we have 5 spaces.
0.1 / 5 = 0.02
So, the value of x is 5.04.
Hence, option (1) is correct.
Problem 6 :
A typist can type 160 words in 5 minutes. At this rate, how long does the typist take to type 320 words ?
(1) 10 minutes (3) 32 minutes 
(2) 2 minutes (4) 64 minutes 
Solution :
160 words = 5 minutes
Multiplying by 2 on both sides
320 words = 10 minutes.
So, a typist can type 320 words in 10 minutes.
Problem 7 :
What is the missing number in the x ?
7 : x = 3 : 15
(1) 19 (3) 24 
(2) 21 (4) 35 
Solution :
By using cross multiplication on we get,
x = 35
So, option (4) is correct.
Problem 8 :
Four letters P, E, A and R are shown below. How many of the letters has/have a line of symmetry ?
(1) 1 (3) 3 
(2) 2 (4) 4 
Solution :
Two letters has a line of symmetry.
So, option (2) is correct.
Problem 9 :
In the figure below, AF, BD and CE are straight lines. ∠AFE = 43º, ∠AFB is a right angle. Find ∠CFD.
(1) 141^{º} (3) 133^{º} 
(2) 137^{º} (4) 129^{º} 
Solution :
By observing the figure,
Given, ∠AFE = 43^{º},
∠AFB is a right angle.
So, ∠AFB = 90^{º}
∠CFD = ∠AFE + ∠AFB
∠CFD = 90^{º} + 43^{º}
∠CFD = 133^{º}
So, option (3) is correct.
Problem 10 :
What is the area of triangle ABC ?
(1) 90 cm^{2} (3) 120 cm^{2} 
(2) 108 cm^{2} (4) 240 cm^{2} 
Solution :
Area of triangle = 1/2 (b × h)
= 1/2 (24 × 10)
= 1/2 × 240
= 120 cm^{2}
So, option (3) is correct.
Problem 11 :
The ratio of the cost of an eraser to the cost of a pen is 2 : 5. A ruler costs twice as much as the eraser. What is the ratio of the cost of the ruler to the cost of the eraser to the total cost of the three items ?
(1) 2 : 4 : 5 (3) 2 ; 4 : 11 
(2) 4 : 2 : 5 (4) 4 : 2 : 11 
Solution :
The cost of an eraser = 2
The cost of a pen = 5
A ruler costs twice as much as the eraser. So, the cost of an eraser = 4
Total cost of the three items = 2 + 5 + 4
= 11
So, the ratio of the cost of the ruler to the cost of the eraser = 4 : 2 : 11.
Hence, option (4) is correct.
Problem 12 :
The average mass of four boys is 52 kg. The masses of Cameron and James are shown in the table below. Leon and Zach are of the same mass each. What is the total mass of Zach and Cameron ?
(1) 55 kg (3) 97 kg 
(2) 92 kg (4) 110 kg 
Solution :
Average mass of four boys is 52 kg.
The masses of Cameron = 42 kg
The masses of James = 56 kg
The Leon and Zach are of the same mass.
So, The masses of Leon = x
The masses of Zach = x
Total mass is
x = 55
Leon and Zach are of the same mass of 55 and 55.
So, option (1) is correct.
Problem 13 :
Which one of the figures below shows that 40% of the figure is shaded ?
Figure A Figure C 
Figure B Figure D 
(1) Figure A (3) Figure C 
(2) Figure B (4) Figure D 
Solution :
Total square box is 20.
40% of the figure is shaded is 40/100.
= 0.4
= 0.4 × 20
= 8
So, option (c) is correct.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM