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EPFO Assistant Numerical Ability Practice Set 1

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EPFO Assistant Numerical Ability Practice Set 1

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EPFO Assistant – Preliminary Examination, conducted in online Mode, has: a duration of 60 Minutes, a total of 100 questions, a maximum score of 100 marks, and consists of 3 sections, namely – Reasoning Ability, English Language, and Numerical Ability/Aptitude.
The article EPFO Assistant Numerical Ability Practice Set 1 presents the EPFO Assistant Numerical Ability section wise questions and answers.

shape Pattern

EPFO Assistant Exam Pattern - Prelims
S. No. Name of the Test (Objective Tests) No. of Questions Maximum Marks Duration Version
1. English Language 30 30 20 Minutes English
2. Reasoning Ability 35 35 20 Minutes Bilingual
3. Numerical Ability 35 35 20 Minutes Bilingual
Total 100 Questions 100 Marks 1 Hour

shape Syllabus

S.No. Topics
1. Missing Numbers
2. Series Completion
3. Profit & Loss %
4. Market price,Cash price & Expenditure problems
5. Fraction
6. Ratios
7. Average & Volume
8. Factoring (LCM, HCF)
9. Mensuration formulas
10. Simple interest & Compound interest

shape Samples

Directions(1-2): What will come in place of question mark (?) in the following series? 1. 320 160 240 600 2100 ?
    A. 8350 B. 8850 C. 9450 D. 9650

Answer: Option C
Explanation:
Series Pattern Given Series
320 320
320 x [latex]\frac{1}{2}[/latex] = 160 160
160 x [latex]\frac{4}{2}[/latex] = 240 240
240 x [latex]\frac{5}{2}[/latex] = 600 600
600 x [latex]\frac{7}{2}[/latex] = 2100 2100
2100 x [latex]\frac{9}{2}[/latex] = 9450 9450

2. 432 443 463 500 562 ?
    A. 659 B. 657 C. 653 D. 651

Answer: Option B
Explanation:
Series Pattern Given Series
432 432
432 + ([latex]{1}^{2}[/latex] + 10) = 443 443
443 + ([latex]{3}^{2}[/latex] + 11) = 463 463
463 + ([latex]{5}^{2}[/latex] + 12) = 500 500
500 + ([latex]{7}^{2}[/latex] + 13) = 562 2100
562 + ([latex]{9}^{2}[/latex] + 14) = 657 657
Directions(1-2): Choose the correct alternative that will continue the same pattern and replace the question mark in the given series.
1. 120, 99, 80, 63, 48, ?
    A. 35 B. 38 C. 39 D. 40

Answer: Option A
Explanation: The pattern is - 21, - 19, - 17, - 15,..... So, missing term = 48 - 13 = 35.
2. In series 2, 6, 18, 54, ...... what will be the [latex]{8}^{th}[/latex] term?
    A. 4370 B. 4374 C. 7443 D. 7434

Answer: Option B
Explanation: Clearly, 2 x 3 = 6, 6 x 3 = 18, 18 x 3 = 54,..... So, the series is a G.P. in which a = 2, r = 3. ∴[latex]{8}^{th}[/latex] term = a[latex]{r}^{8-1}[/latex] = a[latex]{r}^{7}[/latex] = 2 x [latex]{3}^{7}[/latex] = (2 x 2187) = 4374.
1. On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
    A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60

Answer: Option D
Explanation: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls) ⇒ C.P. of 12 balls = S.P. of 17 balls = Rs.720. ⇒ C.P. of 1 ball = Rs.[[latex]\frac{720}{12}[/latex] ] = Rs. 60.
2. The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
    A. Rs. 2000 B. Rs. 2200 C. Rs. 2400 D. Data inadequate

Answer: Option A
Explanation: Let C.P. be Rs. x. Then, [latex]\frac{1920 - x}{x}[/latex] x 100 = [latex]\frac{x - 1280}{x}[/latex] x 100 ⇒1920 - x = x - 1280 ⇒ 2x = 3200 ⇒x = 1600 ∴ Required S.P. = 125% of Rs. 1600 = Rs. [[latex]\frac{125}{100}[/latex] x 1600] = Rs 2000.
1. 5[latex]\frac{1}{5}[/latex] + 4[latex]\frac{1}{2}[/latex] + 4[latex]\frac{1}{3}[/latex] = ?
    A. 14[latex]\frac{1}{10}[/latex] B. 13[latex]\frac{1}{5}[/latex] C. [latex]\frac{13}{30}[/latex] D. 14[latex]\frac{1}{30}[/latex]

Answer: Option D
Explanation: = 5[latex]\frac{1}{5}[/latex] + 4[latex]\frac{1}{2}[/latex] + 4[latex]\frac{1}{3}[/latex] = 5 + 4 + 4 + [latex]\frac{1}{5}[/latex] + [latex]\frac{1}{2}[/latex] + [latex]\frac{1}{3}[/latex] = 13 + [latex]\frac{31}{30}[/latex] = 13 + 1[latex]\frac{1}{30}[/latex] = 13 + 1 + [latex]\frac{1}{30}[/latex] = 14 + [latex]\frac{1}{30}[/latex] = 14 [latex]\frac{1}{30}[/latex]
2. [latex]\frac{4}{7}[/latex] of [latex]\frac{2}{3}[/latex] of [latex]\frac{5}{6}[/latex] of 1008 is
    A. 200 B. 144 C. 64 D. 400

Answer: Option A
Explanation: = [latex]\frac{4}{7}[/latex] ([latex]\frac{2}{3}[/latex]([latex]\frac{5}{6}[/latex]([latex]\frac{5}{8}[/latex] x 1008))) = [latex]\frac{4}{7}[/latex] x [latex]\frac{2}{3}[/latex] x [latex]\frac{5}{6}[/latex] x [latex]\frac{5}{8}[/latex] x 1008 = 200
1. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?
    A. Rs. 500 B. Rs. 1500 C. Rs. 2000 D. None of these

Answer: Option C
Explanation: Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively. Then, 4x - 3x = 1000 ⇒ x = 1000. ∴ B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.
2. The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?
    A. 3 : 3 : 10 B. 3 : 3 : 10 C. 23 : 33 : 60 D. Cannot be determined

Answer: Option C
Explanation: Let A = 2k, B = 3k and C = 5k. A's new salary = [latex]\frac{115}{100}[/latex] of 2k = [[latex]\frac{115}{100}[/latex] x 2k] = [latex]\frac{23k}{10}[/latex] B's new salary = [latex]\frac{110}{100}[/latex] of 3k = [[latex]\frac{110}{100}[/latex] x 3k] = [latex]\frac{33k}{10}[/latex] C's new salary = [latex]\frac{120}{100}[/latex] of 5k = [[latex]\frac{120}{100}[/latex] x 5K] = 6k ∴ New ratio [[latex]\frac{23k}{10}[/latex]: [latex]\frac{33k}{10}[/latex]: 6k] = 23 : 33 : 60

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