Data Interpretation SSC BANK pdf (chsl-clerk-po exem)

Data Interpretation SSC BANK pdf (chsl-clerk-po exem)

Here i am sharing with you question on  Data Interpretation SSC BANK pdf (chsl-clerk-po exem). In any exam reasoning section carries great significance. in SSC CHSL and SSC CGL exam important topic is Data Interpretation little bit more practice. in this topic you can more score in exam.

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download Data Interpretation SSC BANK pdf (chsl-clerk-po exem)

most expected question for Data Interpretation SSC BANK pdf (chsl-clerk-po exem)

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Permutation & Combination Problems with Solutions in pdf Bankers Way

simple and compound interest problems with solutions PDF bankers way

Data Interpretation Questions with solutions

permutation & combination questions

1. At an election, three wards of a town are to be canvassed by 4, 5 and 8 men respectively. If there are
20 men, in how many ways they can be allotted to the different wards?
(a) 4!*5!*8!/20! (b) 1240 (c)20890 (d)20!/17! (e)not
2. In how many different ways can 4 boys and 3 girls be arranged in a row such that all boys stand
together and all girls stand together?
(a) 288 (b) 144 (c) 56 (d) 64 (e) NOT
3. Out of 5 men and 3 women, a committee of 3 members is to be formed so that it has 1 woman and 2
men. In how many different ways can it be done?
(a) 24 (b)30 (c) 45 (d) 72 (e) 10
4. In how many different ways can the letters of the word LONELY be arranged, so that the vowels are
at the two ends?
(a) 120 (b) 720 (c) 360 (d) 48 (e) 60
5. While packing for a business trip Rajesh has packed 3 pairs of shoes, 4 pants, 3 half-pants, 6 shirts, 3
sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of ‗lower wear‘, a
choice of ‗upper wear‘ and finally he may or may not choose to wear a jacket. How many different outfits
are possible?
(a) 240 (b) 3! (c) 5*3! (d) 2!*3! (e) 378
6. The number of ways in which the letters of the word CAPSULE can be rearranged so that the even
places are always occupied by consonants is:
(a) 576 (b) 6! (c) 4!*2! (d) 3!*6 (e) NOT
7. The number of ways in which 20 different flowers of two colors can be set alternatively on a
necklace, there being 10 flowers of each color is:
(a) 20! (b)10! * 10! (c) 20!*10 (d) 10!*2 (e) NOT

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