INDIAN NAVY ENTRANCE TEST REASONING NUMERICAL ABILITY MCQ
INDIAN NAVY ENTRANCE TEST REASONING NUMERICAL ABILITY MCQ
Question – 1 : Direction: Find the missing number in the given question. 7 22 5 5 9 11 9 11 15 38 77 ?
Answer – (A) : 160
Answer – (B) : 120
Answer – (C) : 83
Answer – (D) : 55
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Correct Answer : (A)
Explanation : In the first column, (9×5) -7 = 38In the second column, (11×9) – 22 = 77Similarily in third column, (11×15) – 5 = 160
Question – 2 : 1.27 is equal to:
Answer – (A) : 1411
Answer – (B) : 73100
Answer – (C) : 1411
Answer – (D) : 1114
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Correct Answer : (C)
Explanation : In the given recurring decimal, the non-recurring digit is 1, recurring digits are 27.Let x = 1.27¯ = 1.272727….. (i)100x = 127.272727….. (ii)Subtraction eqution (i) from (ii) we’ll get99x = 126⇒ x = 12699=14111.27 is equal to 1411
Question – 3 : Two articles are bought for ₹ 2,600 each. One of them is sold at a loss of 4% and the average selling price of both the articles is ₹ 2,652. The other article is sold at a profit percent or loss percent of
Answer – (A) : profit 8%
Answer – (B) : loss 6%
Answer – (C) : profit 8%
Answer – (D) : profit 6%
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Correct Answer : (C)
Explanation : For 1st article,C.P.1 = 2600It was sold at a loss of 4%Loss =C.P.100×4=2600100×4Loss = 104S.P.1 = Rs 2496Now,Average S.P. =S.P.1+S.P.22S.P.2 = Average S.P. ×2− S.P.1S.P.2 =2652×2−2496S.P.2 = Rs 2808For 2nd Article,C.P.2 = Rs 2600S.P.2 = Rs 2808Since C.P.2 is less than S.P.2 , there is profit.Profit =(S.P.−C.P.)C.P.×100=(2808−2600)2600×100Profit =8%
Question – 4 : The compound interest is Rs. 840 for 2 years at 10 percent rate of interest then find the principal amount?
Answer – (A) : Rs. 4000
Answer – (B) : Rs. 4500
Answer – (C) : Rs. 3500
Answer – (D) : Rs. 4100
View Answer
Correct Answer : (A)
Explanation : Given:Compound interest (C.I) = Rs. 840Time = 2 yearsRate of interest = 10 percentFormula:C.I = P × 1+r100n-1Here,P = Principal Amount, r = Rate of interest, n = Time in yearsAccording to the question,840 = P × 1+101002–1⇒ 840 = P × 1+1102–1⇒ 840 = P × 11102–1⇒ 840 = P × 121100–1⇒ 840 = P × 121-100100⇒ 840 = P × 21100⇒ P = 4000∴ The Principal Amount will be Rs. 4000.
Question – 5 : A sphere of radius 3 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 6 cm. If the sphere is submerged completely. Then the surface of the water is raised by –
Answer – (A) : 1 cm
Answer – (B) : 2 cm
Answer – (C) : 3 cm
Answer – (D) : 4 cm
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Correct Answer : (A)
Explanation : Volume of cylinder = πr2HVolume of sphere = 43 πR3Where,r → Radius of the cylinderH → Height of the cylinderR → Radius of the sphereWhen the sphere is submerged then,The volume of sphere = Volume of cylinder⇒ 43 × π × (3)3 = π × (6)2 × H⇒ H = 3636⇒ H = 1 cm∴ The surface of the water is raised by 1cm
Question – 6 : Find the difference between CI & SI for 2 years at a 20% rate of interest. If the principal is 800 rupee.
Answer – (A) : 32
Answer – (B) : 20
Answer – (C) : 42
Answer – (D) : 0
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Correct Answer : (A)
Explanation : Given:The principal is 800 rupee.Rate is 20%.Formula:SI = P×R×T100CI = P1+r100n-1Here, n is the number of terms in the year.The simple interest for 2 years.= 800×20×2100= 8 × 20 × 2= 320 rupeeNow, Compound interest for 2 years is:= 8001+201002-1= 8001201002-1= 80012102-1= 800 × 144-100100= 352 rupeeNow, The difference between CI and SI is:= 352 – 320= 32∴ The difference between CI and SI is 32.
Question – 7 : In a certain code, “TABLE” is written as “150”. How will “DESK” be written in that code?
Answer – (A) : 120
Answer – (B) : 80
Answer – (C) : 120
Answer – (D) : 60
View Answer
Correct Answer : (C)
Explanation : The logic is:Number of letters in the word ‘TABLE’ is 5 and 30 × 5 = 150Similarly,Number of letters in the word ‘DESK’ is 4 and 30 × 4 = 120So, ‘120’ is the correct answer.
Question – 8 : X can do a piece of work in 30 days. X leaves after working for 3 days and the remaining work is completed by Y in 18 days. How much time is taken by Y to complete the piece of work alone?
Answer – (A) : 20 days
Answer – (B) : 15 days
Answer – (C) : 10 days
Answer – (D) : 20 days
View Answer
Correct Answer : (D)
Explanation : Given:X can do a piece of work in =30 days,X leaves after working for =3 days,Y finished the remaining work in =18 days, Calculation:X′ s 1 day’s work =130X ‘s 3 day’s work =330⇒110∴ remaining work =1−110⇒910Now, 910 work is done in 18 days∴1 piece of work is done by Y in =109×18⇒10×2⇒20 days∴Y can complete the piece of work alone in 20 days.
Question – 9 : Find the missing number. 48596863625612?
Answer – (A) : 10
Answer – (B) : 15
Answer – (C) : 14
Answer – (D) : 10
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Correct Answer : (D)
Explanation : 4×9×63=2163=68×6×363=17283=125×8×253=10003=10
Question – 10 : Direction: The sheet of paper shown in the figure (x) given on the left hand side, in each problem, is folded to form a box. Choose from amongst the alternatives (A), (B), (C) and (D), the boxes that are similar to the box that will be formed.Select the dice that can be formed by folding the given sheet along the lines.
Answer – (A) : Only (A) and (B)
Answer – (B) : Only (B) and (C)
Answer – (C) : Only (A) and (B)
Answer – (D) : Only (C)
View Answer
Correct Answer : (C)
Explanation : When the dice that can be formed by folding the given sheet along the lines, the following numbers will be opposite to each other is shown below:1 is opposite to 6.2 is opposite to 5.3 is opposite to 4.So, the dice (A) and (B) are the only possible dice formed after folding.
Question – 11 : Find the missing term: 3, 4, 7, 7, 13, 13, 21, 22, 31, 34, ?
Answer – (A) : 43
Answer – (B) : 43
Answer – (C) : 51
Answer – (D) : 52
View Answer
Correct Answer : (B)
Explanation : The given sequence is a combination of two series:I. 3, 7, 13, 21, 31, ? andII. 4, 7, 13, 22, 34The pattern in I is + 4, + 6, + 8, + 10, ……The pattern in II is + 3, + 6, + 9, + 12, …………..So, missing term = 31 + 12 = 43
Question – 12 : If the sixth term of an A.P. is zero, then t18t9 is ?Where tn denotes the nth term of AP.
Answer – (A) : 4
Answer – (B) : 4
Answer – (C) : 5
Answer – (D) : 6
View Answer
Correct Answer : (B)
Explanation : Given,The sixth term of an A.P. is zeroWe know that,Sum of the first n terms in AP=n2(a+l)Where, a= First term, d= Common difference, n= number of terms, an=nth term and l= Last termLet the first term of AP be ‘a’ and the common difference be ‘d’Therefore,a6=0⇒a+(6−1)×d=0⇒a+5d=0⇒a=−5dTo Find, t18t9t18t9 =a+17 da+8 d⇒t18t9 =−5 d+17 d−5 d+8 d⇒t18t9 =12 d3 d⇒t18t9 =4
Question – 13 : A shopkeeper sells his goods at 25% profit. If he had bought it at 5% less and sold it for Rs. 55 less, he would have gained 20%. Find the original selling price of the article?
Answer – (A) : Rs. 625
Answer – (B) : Rs. 600
Answer – (C) : Rs. 625
Answer – (D) : Rs. 575
View Answer
Correct Answer : (C)
Explanation : Given:Original Profit% = 25%Change in CP = 5% less than Original CPOriginal SP – New SP = Rs. 55New Profit % = 20%Concept:Consider the CP be 100y, then proceed.Let, the Original CP = 100yOriginal SP = 125yNew CP = 95y∵ Profit Made on New CP = 20% × 95y= 19y∴ New SP = CP + Profit= 95y + 19y= 114y∵ Original SP – New SP = Rs. 55⇒ 125y – 114y = 55⇒ 11y = 55⇒ y = 5511⇒ y = 5∴ Original SP = 125y= 125 × 5= Rs. 625
Question – 14 : Find the missing term: 9, 27, 31, 155, 161, 1127, ?
Answer – (A) : 1135
Answer – (B) : 1135
Answer – (C) : 1288
Answer – (D) : 2254
View Answer
Correct Answer : (B)
Explanation : 9×3=2727+4=3131×5=155155+6=161161×7=11271127+8=1135So, the missing term =1135
Question – 15 : If 107(1−2.43×10−3) = 1.417 + x, then x is equal to:
Answer – (A) : 0.0081
Answer – (B) : 0.417
Answer – (C) : 0.0081
Answer – (D) : 0.81
View Answer
Correct Answer : (C)
Explanation : Given:107(1−2.43×10−3) = 1.417 + x⇒ 107(1−0.00243) = 1.417 + x⇒ 107 × 0.99757 = 1.417 + x⇒ 1.4251 = 1.417 + x⇒ x = 1.4251 – 1.417∴ x = 0.0081
Question – 16 : What is the third proportional to 16 and 24 ?
Answer – (A) : 36
Answer – (B) : 34
Answer – (C) : 32
Answer – (D) : 36
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Correct Answer : (D)
Explanation : The third proportional proportion is the second term of the mean terms.For example, if we have a:b=c:d , then the term ‘ c ‘ is the third proportional to ‘ a ‘ and ‘ b ‘Represented as:a:b::b:cCalculation:Let the third proportion to 16 and 24 be x⇒1624=24x⇒x=(24×24)16⇒x=36∴ The third proportional to 16 and 24 is 36
Question – 17 : In a certain code language NOSE is coded as 10 and SEAO is coded as 8. How GERM coded in that code language?
Answer – (A) : 8
Answer – (B) : 8
Answer – (C) : 9
Answer – (D) : 10
View Answer
Correct Answer : (B)
Explanation : As per the English alphabetical series and its place value. The logic followed here is: N → 14 O → 15 S → 19 E → 5 NOSE → 14 + 15 + 19 + 5 = 53 = (5 + 3) + (5 – 3) = (8 + 2) = 10 And- S → 19 E → 5 A → 1 O → 15 SEAO → 19 + 5 + 1 + 15 = 40 = (4 + 0) + (4 – 0) = (4 + 4) = 8 So, G → 7 E → 5 R → 18 M → 13 GERM → 7 + 5 + 18 + 13 → 43 = (4 + 3) + (4 – 3) = (7 + 1) = 8 So, code for GERM is 8.
Question – 18 : Travelling at 34 of the normal speed, a person reaches his workplace 15 minutes late. How many minutes does he take usually to reach the workplace?
Answer – (A) : 45 minutes
Answer – (B) : 30 minutes
Answer – (C) : 42 minutes
Answer – (D) : 45 minutes
View Answer
Correct Answer : (D)
Explanation : Let the normal speed be 4 km/hrTravelling speed = 4 × 34 = 3 km/hrRatio of normal speed to travelling speed = 4x : 3xAs we know,Time is inversely proportional to time.Ratio of normal time to travelling time = 3x : 4x4x – 3x = 15⇒ x = 15∴ Normal time = 3x = 3 × 15 = 45 min
Question – 19 : A can complete a work in 14 days while B can do it in 20 days. They started working together but after some days. A Left the work and the whole work will complete in 10 days, Find after how many days A left the work?
Answer – (A) : 7 days
Answer – (B) : 9 days
Answer – (C) : 7 days
Answer – (D) : 5 days
View Answer
Correct Answer : (C)
Explanation : Given:A can complete the work in 14 days.B can complete the work in 20 days.A + B together work for some days and then A leftTotal work completed in 10 days.Total work = LCM(14, 20) = 140 unit⇒ One day’s work of A = 14014 = 10⇒ One day’s work of B = 14020 = 7Let A and B worked together for X days, and then A left⇒ B alone worked for (10 – X) daysTotal work = ( (10 + 7) × X) + (7 × (10 – X) ) = 140⇒ (17 × X) + 70 – (7 × X) = 140⇒ (17 × X) – (7 × X) = 140 – 70⇒ 10 × X = 70⇒ X = 7⇒ A and B worked together for 7 days and then A left∴ A left after 7 days.
Question – 20 : The average age of the boys in a class of 20 boys is 15.6 years. What will be the average age if 5 new boys come whose average is 15.4 years?
Answer – (A) : 15.56 yrs
Answer – (B) : 15.65 yrs
Answer – (C) : 18 yrs
Answer – (D) : 16.56 yrs
View Answer
Correct Answer : (A)
Explanation : We have, Average age = Sum of all agesNo. of persons∴ 15.6 = Sum of 20 boys ages20⇒ Sum of ages of 20 boys = 20 × 15.6 = 312After 5 new boys come, total boys = 20 + 5 = 25Sum of ages of 5 new boys = 15.4 × 5 = 77Sum of 25 boys ages = 312 + 77 = 389Overall average = 38925 = 15.56 yrs.
Question – 21 : In the question, select the related word from the given alternatives.Country : President :: State:?
Answer – (A) : Governor
Answer – (B) : Prime Minister
Answer – (C) : Speaker
Answer – (D) : Governor
View Answer
Correct Answer : (D)
Explanation : President is the highest officer of Country and Governor is the highest officer of State.
Question – 22 : Given the ratio of a : b = 5 : 7, b : c = 21 : 6 then find the value of (3a + c) : (2b – c).
Answer – (A) : 17:12
Answer – (B) : 12:17
Answer – (C) : 17:12
Answer – (D) : 05:03
View Answer
Correct Answer : (C)
Explanation : Given:a : b = 5 : 7b : c = 21 : 6Concept:Make the ratio of common entity equal.After multiplying 3 in both denominator and numerator of first ratio we get:a : b = 15 : 21And b : c = 21 : 6So, a : b : c = 15 : 21 : 6So we have to find the value of (3a + c) : (2b – c) = (3 × 15 + 6) : (2 × 21 – 6)∴ (3a + c) : (2b – c) = 51 : 36 = 17 : 12
Question – 23 : An Rs 100 shirt is offered at 10% discount and a Rs 300 pair of trousers at 20% discount. If Pritam bought 1 shirt and 3 pairs of trousers, what is the effective discount (in %) he got?
Answer – (A) : 19
Answer – (B) : 18
Answer – (C) : 17
Answer – (D) : 16
View Answer
Correct Answer : (A)
Explanation : Marked price of shirt = Rs 100discount =10% of marked price=10100×100= Rs 10 . Marked price of trouser = Rs 300 . discount on trouser =20% of marked price=20100×300= Rs 60 . The Marked price of 3 trouser = Rs 900So, total discount on trouser = Rs 180total Marked price =(900+100)=1000Hence, total Discount =(180+10)So, discount %=1901000×100=19%Therefore, the effective discount will be 19% .
Question – 24 : Find how many odd factors are there in the number 1240.
Answer – (A) : 4
Answer – (B) : 4
Answer – (C) : 5
Answer – (D) : 16
View Answer
Correct Answer : (B)
Explanation : Given number = 1240 Odd factors of pa×qb=(a+1)(b+1) Where p and q should be odd prime numbers Factor of 1240=23×51×(31)1 ⇒ Odd number are 51×(31)1 ⇒ Odd factors of pa×qb ⇒(a+1)(b+1) ⇒(1+1)(1+1)=4 ∴1240 has 4 odd factors.
Question – 25 : Unscramble the word: OPSOESRCR
Answer – (A) : PROCESSOR
Answer – (B) : PROCESSOR
Answer – (C) : PROSSECOR
Answer – (D) : PORCESSOR
View Answer
Correct Answer : (B)
Explanation : The correct word is PROCESSOR.A processor is an integrated electronic circuit that performs the calculations that run a computer. A processor performs arithmetical, logical, input/output (I/O) and other basic instructions that are passed from an operating system (OS). Most other processes are dependent on the operations of a processor.