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TSCAB Recruitment Quantitative Aptitude – Staff Assistant 2019

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TSCAB Recruitment Quantitative Aptitude – Staff Assistant 2019

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Telangana State Cooperative Apex Bank has announced a notification for the recruitment of Staff Asst Recruitment posts. The Candidates who are interested and meet all eligibility criteria can read the Official Notification of TSCAB Staff Asst can apply through the Official website or from the given link. Get the complete details of TSCAB Recruitment Quantitative Aptitude Syllabus along with exam pattern and samples. Check the most important questions related to TSCAB Recruitment Quantitative Aptitude section. Candidates can score maximum marks of Quantitative Aptitude section of TSCAB Recruitment.

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TSCAB Recruitment Quantitative Aptitude Exam Pattern
S. No. Name of Tests (objective) No. of questions Max. Marks Time allotted for each test
(Separately timed)
1. General/Financial Awareness 50 50 35 Minutes
2. General English 40 40 35 Minutes
3. Reasoning Ability & Computer Aptitude. 50 50 45 Minutes
4. Quantitative Aptitude 50 50 45 Minutes
Total 190 190 160 Minutes


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TSCAB Recruitment TSCAB Recruitment Quantitative Aptitude – Syllabus (50 Marks):

TSCAB Recruitment Quantitative Aptitude – Syllabus
1. Number system
2. Volume and Surface Area
3. Simplification
4. Permutation and Combination
5. Time and Distance
6. Partnership
7. Logarithm
8. Time & Work
9. Functions
10. Simple Interest
11. Problems on H.C.F and L.C.M.
12. Mensuration


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1. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?


    A. Rs. 4462.50
    B. Rs. 8032.50
    C. Rs. 8900
    D. Rs. 8925
    E. None of these


Answer: Option D


Explanation:

Principal
= Rs. \(\frac{100 \times 4016.25}{9 \times 5}\)

= Rs. \(\frac{401625}{45}\)

= Rs. 8925.

1. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:


    A. 720
    B. 900
    C. 1200
    D. 1800


Answer: Option C


Explanation:

2(15 + 12) x h = 2(15 x 12)

h = \(\frac{180}{27}\) m = \(\frac{20}{3}\) m.
Volume = (\({15 \times 12 \times \frac {20}{3}}\)) m\(^{3}\) = 1200 m\(^{3}\).

1. If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, then x is equal to:


    A. 1
    B. 3
    C. 5
    D. 10


Answer: Option B


Explanation:

\(log_{10}\) 5 + \(log_{10}\) (5x + 1) = \(log_{10}\) (x + 5) + 1

\(log_{10}\) 5 + \(log_{10}\) (5x + 1) = \(log_{10}\) (x + 5) + \(log_{10}\) 10

\(log_{10}\) [5 (5x + 1)] = \(log_{10}\) [10(x + 5)]

5(5x + 1) = 10(x + 5)

5x + 1 = 2x + 10

3x = 9

x = 3.

1. A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:


    A. 30 birds
    B. 60 birds
    C. 72 birds
    D. 90 birds


Answer: Option A


Explanation:

Let the total number of shots be x. Then,

Shots fired by A = \(\frac {5}{8}\)x

Shots fired by B = \(\frac {3}{8}\)x

Killing shots by A = \(\frac {1}{3}\) of \(\frac {5}{8}\) x = \(\frac {5}{24}\)x

Shots missed by B = \(\frac {1}{2}\) of \(\frac {3}{8}\) x = \(\frac {3}{16}\)x

\(\frac {3 \times x}{16}\) = 27 or x = \(\frac {27 \times 16}{3}\) = 144.

Birds killed by A = \(\frac {5x}{24}\) = \(\frac {5}{24}\) x 144 = 30.

1. A farmer traveled 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 traveled on foot is:


    A. 14 km
    B. 15 km
    C. 16 km
    D. 17 km


Answer: Option C


Explanation:

Let the distance traveled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, \(\frac {x}{4}\) + \(\frac {(61 -x)}{9}\) = 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

1. Three flags each of different colors are available for a military exercise, Using these flags different codes can be generated by waving
I. Single flag of different colors
II. Any two flags in a different sequence of colors.
III. three flags in a different sequence of colors.
The maximum number of codes that can be generated is.


    A. 6
    B. 9
    C. 15
    D. 18


Answer: Option C

Explanation:
This type of question becomes very easy when we assume three colors are red(R) blue(B) and Green(G).
We can choose any color.
Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, the number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colors, the number of ways is six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15.

1. Rohan and Mohan invest in the ratio 5: 2. 10% of the profit is donated to a hospital before dividing it between the two. Rohan gets Rs. 6840. What is Mohan’s share?

    A. Rs. 2736
    B. Rs. 3800
    C. Rs. 4788
    D. Rs. 6840


Answer: Option A

Explanation:
Ratio of investment = Ratio of Profit
Ratio of Profit of Rohan to Mohan = 5:2
Rohan’s Share = Rs. 6840 = \(\frac {5}{5+2}\) x (90% Total Profit) ———-> 10% given to hospital

∴ 6840 = \(\frac {5}{7}\) x \(\frac {90}{100}\) x Total Profit

∴ Total Profit = Rs. 10,640/-
Vijay’s Share = 10,640 – 6840 – 1064 = Rs. 2736/-

1. A drink vendor has 80 liters of Maaza, 144 liters of Pepsi and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn’t want to mix any two drinks in a can. What is the least number of cans required?

    A. 35
    B. 37
    C. 42
    D. 30


Answer: Option B

Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = \(\frac{80}{16}\) = 5
Number of cans of Pepsi = \(\frac{11}{16}\) = 9
Number of cans of Sprite = \(\frac{368}{16}\) = 23
The total number of cans required = 5 + 9 + 23 = 37 cans.

TSCAB – Related Information
TSCAB Recruitment Notification
TSCAB Recruitment Eligibility
TSCAB Recruitment Exam Pattern