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JEE Exam  >  JEE Tests  >  SRMJEEE Subject Wise & Full Length Mock Tests 2025  >  SRMJEE Aptitude Mock Test - 1 - JEE MCQ

SRMJEE Aptitude Mock Test - 1 - JEE MCQ


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20 Questions MCQ Test SRMJEEE Subject Wise & Full Length Mock Tests 2025 - SRMJEE Aptitude Mock Test - 1

SRMJEE Aptitude Mock Test - 1 for JEE 2025 is part of SRMJEEE Subject Wise & Full Length Mock Tests 2025 preparation. The SRMJEE Aptitude Mock Test - 1 questions and answers have been prepared according to the JEE exam syllabus.The SRMJEE Aptitude Mock Test - 1 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SRMJEE Aptitude Mock Test - 1 below.
Solutions of SRMJEE Aptitude Mock Test - 1 questions in English are available as part of our SRMJEEE Subject Wise & Full Length Mock Tests 2025 for JEE & SRMJEE Aptitude Mock Test - 1 solutions in Hindi for SRMJEEE Subject Wise & Full Length Mock Tests 2025 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt SRMJEE Aptitude Mock Test - 1 | 20 questions in 25 minutes | Mock test for JEE preparation | Free important questions MCQ to study SRMJEEE Subject Wise & Full Length Mock Tests 2025 for JEE Exam | Download free PDF with solutions
SRMJEE Aptitude Mock Test - 1 - Question 1
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What is the area (in sq. cm) of a regular hexagon of side 9 cm?

Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 1

Area of a hexagon =
Here, side of the hexagon (a) = 9 cm
Thus,
Area of the hexagon (A) =

SRMJEE Aptitude Mock Test - 1 - Question 2
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The sum of twice a fraction and 5 times its reciprocal is 7. What is the fraction?

Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 2
Let the fraction be 1/x.

5x2 + 2 - 7x = 0
5x2 - 5x - 2x + 2 = 0
5x(x - 1) - 2(x - 1) = 0
(x - 1)(5x - 2) = 0
x = 1, 2/5
Thus, the required fraction is 5/2.
This is correct answer.
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SRMJEE Aptitude Mock Test - 1 - Question 3
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Let O be the centre of a circle and AC be its diameter. BD is a chord intersecting AC at E. Point A is joined to B and D. If ∠BOC = 50° and ∠AOD = 110°, then ∠BEC = ?

Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 3


We know that the sum of angles on a straight line is 180°.
∠AOD + ∠EOD = 180°
∠EOD = 70°
∴ ∠BOD = 120°
We know, OB = OD = Radius of the circle
Here, ∠OBE = ∠ODE
Now, in ΔBOD,
120° + 2x = 180°
2x = 180° - 120°
x = 30°
Now, in ΔBOE,
50° + 30° + ∠BEO = 180°
∠BEO = 100°
We know that the sum of angles on a straight line is 180°.
∠BEO + ∠BEC = 180°
∠BEC = 80°

SRMJEE Aptitude Mock Test - 1 - Question 4
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lf a + b + c = 11 and ab + bc + ca = 38, then a3 + b3 + c3 - 3abc is equal to:
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 4
Given: a + b + c = 11 and ab + bc + ac = 38
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca)
a2 + b2 + c2 = (11)2 - 2(38)
a2 + b2 + c2 = 121 - 76
a2 + b2 + c2 = 45
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
a3 + b3 + c3 - 3abc = (11){45 - (ab + bc + ca)}
a3 + b3 + c3 - 3abc = (11)(45 - 38)
a3 + b3 + c3 - 3abc = 11 × 7
a3 + b3 + c3 - 3abc= 77
SRMJEE Aptitude Mock Test - 1 - Question 5
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The ratio of the total production of type A cars in 2015 and type B cars in 2014 taken together to the production of type C cars in 2017 and type E cars in 2018 taken together is
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 5

Total production of type A cars in 2015 = 56
Total production of type B cars in 2014 = 48
Total production of type A and type B cars in 2015 and 2014, respectively = 56 + 48 = 104
Total production of type C cars in 2017 = 57
Total production of type E cars in 2018 = 60
Total production of type C and type E cars in 2017 and 2018, respectively = 57 + 60 = 117
The ratio of the total production of type A cars in 2015 and type B cars in 2014 taken together to the production of type C cars in 2017 and type E cars in 2018 taken together
= 104 : 117
= 8 : 9

SRMJEE Aptitude Mock Test - 1 - Question 6
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In the data related to the production of type D cars is represented by a pie chart, then the central angle of the sector representing the production of cars in 2015 will be:
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 6

Total number of cars of type D in 2015 = 200
The central angle of the sector representing production of cars in 2015 will be:

So, the central angle is 81° .

SRMJEE Aptitude Mock Test - 1 - Question 7
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The total production of type D cars during 2015-2017 is what percent less than the total production of type E cars during 2014, 2015, 2016 and 2018 taken together?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 7

Total number of cars of type D from 2015 to 2017 = 140
Total number of cars of type E during 2014, 2015, 2016 and 2018 taken together = 200
Cars of type E - Cars of type D = 200 - 140 = 60 cars
Required percentage = = 30%

SRMJEE Aptitude Mock Test - 1 - Question 8
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The total production of type C cars in 2015 and type E cars in 2018 taken together is what percent of the total production of cars in 2014 and 2017 taken together?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 8

According to question:
Total production of type C cars in 2015 = 42
Total production of type E cars in 2018 = 60
So, total number of cars = 42 + 60 = 102 cars
Total production of cars in 2014 = 210 cars
Total production of cars in 2017 = 300 cars
Total production of cars in 2014 and 2017 = 510 cars
Percentage of cars production of type C and E = (102/510) x 100 = 20%
So, the answer is 20%.

SRMJEE Aptitude Mock Test - 1 - Question 9
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A is 40% more efficient than B and C is 20% less efficient than B. Working together, they can complete a task in 20 hours. In how many hours will A alone complete 35% of that task?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 9
Let the efficiency of B be x.
Efficiency of A = 1.4x
Efficiency of C = 0.8x
ATQ,
20(x + 1.4x + 0.8x) = 1
x = 1/64
Efficiency of A = 1.4/64 = 7/320
Hours to complete full task = 320/7
Hours to complete 35% task = 0.35 × (320/7) = 16
SRMJEE Aptitude Mock Test - 1 - Question 10
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Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 10


On simplifying the above equation, we get

SRMJEE Aptitude Mock Test - 1 - Question 11
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The radii of the two circular faces of the frustum of a cone are 5 cm and 4 cm. If the height of the frustum is 21 cm, what is its volume in cm3?  (π = 22/7)
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 11

Volume of the frustum = ; where R and r are the radii of two bases
=
= 22(25 + 16 + 20)
= 22 × 61
= 1342 cm3

SRMJEE Aptitude Mock Test - 1 - Question 12
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The volumes of spheres A and B are in the ratio 125 : 64. If the sum of radii of A and B is 36 cm, then the surface area (in cm2) of A is:
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 12

Ratio of volumes of spheres A and B = 125 : 64


4ra = 5rb - (1)
(ra + rb = 36 cm)
Put the value of rb in eq. (1).
4ra = 5(36 cm - ra)
4ra + 5ra = 180 cm
ra = 180 cm/9 = 20 cm
Thus, surface area = 4πra2 = 1600π cm2
Hence, answer option 3 is correct.

SRMJEE Aptitude Mock Test - 1 - Question 13
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In ΔABC, AB = AC. A circle drawn through B touches AC at D and intersect AB at P. If D is the midpoint of AC and AP = 2.5 cm, then AB is equal to:
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 13


Given: D is the midpoint of AC.
So,
AD = AC/2
But, AC = AB (given)
AD = AB/2 --- (1)
AD is a tangent and APB is a secant. So, the tangent secant theorem can be applied.
AD2 = AP × AB
(AB/2)2= 2.5 cm × AB
AB = 10 cm
Hence, answer option 4 is correct.

SRMJEE Aptitude Mock Test - 1 - Question 14
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Find the value of tan 4° tan 43° tan 47° tan 86°.
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 14

tan 86° = cot (90° - 86°) = cot 4°
tan 47° = cot (90° - 47°) = cot 43°
∴ (tan 4° . tan 86°) (tan 43° . tan 47°)
= (tan 4° . cot 4°) (tan 43° . cot 43°)
= 1 (∵ tanθ . cotθ = 1)

SRMJEE Aptitude Mock Test - 1 - Question 15
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A train starts from station A and reaches station B 15 minutes late when it moves at a speed of 40 km/hr, and 24 minutes late when it moves at a speed of 30 km/hr. The distance between the two stations is
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 15
Distance between stations A and B
=
= = 18 km
SRMJEE Aptitude Mock Test - 1 - Question 16
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The expression (a + b - c)3 + (a - b + c)3 - 8a3 is equal to:
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 16
Let (a + b - c) = A, (a - b + c) = B and -2a = C
(a + b - c)3 + (a - b + c)3 - 8a3
= (A)3 + (B)3 - (C)3
As A + B + C = a + b - c + a - b + c - 2a = 0
(A3 + B3 + C3 = 3ABC when A + B + C = 0)
A3 + B3 + C3 = 3ABC
= 3(a + b - c) (a - b + c) (-2a)
= 6a(a - b + c) (c - a - b)
SRMJEE Aptitude Mock Test - 1 - Question 17
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From the following table, how many patients were in the age group 40-60?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 17
Required number = 45 - 29 = 16
Hence, option (4) is correct.
SRMJEE Aptitude Mock Test - 1 - Question 18
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What is the area (in sq. cm) of a rectangle if its diagonal is 25 cm and one of the sides is 24 cm?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 18
Diagonal of the rectangle = 25 cm
One side = 24 cm
Other side = = = 7 cm
So, area of the rectangle = 24 × 7 = 168 sq. cm
SRMJEE Aptitude Mock Test - 1 - Question 19
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6 times a fraction is greater than 7 times its reciprocal by 11. What is the fraction?
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 19

Let the fraction be1/x.
6(1/x) - 7(x/1) = 11
6 - 7x2 = 11x
7x2 + 11x - 6 = 0
7x2 + 14x - 3x - 6 = 0
7x(x + 2) - 3(x + 2) = 0
⇒ (7x - 3)(x + 2) = 0
Thus,
Either 7x - 3 = 0
⇒ 7x = 3
⇒ x = 3/7
Or x + 2 = 0
x = -2
The fraction is 1/x.
So, the fraction is either 7/3 or -1/2.

SRMJEE Aptitude Mock Test - 1 - Question 20
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An article was sold at a gain of 16%. If it had been sold for Rs. 36 more, the gain would have been 20%. The cost price of the article is
Detailed Solution for SRMJEE Aptitude Mock Test - 1 - Question 20
Let the cost price of an article be Rs. x.
Gain = 16%
Selling price = Rs. x × 116/100 = Rs. 1.16x
Selling price when sold for 20% gain = Rs. x × 120/100 = Rs. 1.2x
According to question,
1.2x - 1.16x = 36
0.04x = 36
x = 900
Thus, the cost price of the article is Rs. 900.
Hence, answer option 2 is correct.
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