INDIAN NAVY ENTRANCE TEST REASONING NUMERICAL ABILITY QUESTIONS AND ANSWERS
INDIAN NAVY ENTRANCE TEST REASONING NUMERICAL ABILITY QUESTIONS AND ANSWERS
Question – 1 : What is the pq state of 0.163― ?
Answer – (A) : 163999
Answer – (B) : 171990
Answer – (C) : 163990
Answer – (D) : 184999
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Correct Answer : (A)
Explanation : Given:0.163―Let the value of pq = x⇒ x = 0.163163163……… —-(1)Multiply by 10 in the equation (1)⇒ 10x = 1.63163163…… —-(2)Multiply by 10 in the equation (2)⇒ 100x = 16.3163163……. —-(3)Multiply by the 10 in the equation (3)⇒ 1000x = 163.163163163…… —-(4)Subtract equation (1) from equation (4)⇒ 999x = 163⇒ x = 163999
Question – 2 : Choose the box that is similar to the box formed from the given sheet of paper (x).
Answer – (A) : (1) and (3) only
Answer – (B) : (2) and (4) only
Answer – (C) : (3) and (4) only
Answer – (D) : (1) and (4) only
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Correct Answer : (A)
Explanation : The figure (x) is similar to Figure (1). So, when the sheet shown in figure (x) is folded to form a box (cuboid), then the two rectangular-shaded faces lie opposite to each other, two rectangular white faces lie opposite to each other and the two square shaped faces (one shaded and one white) lie opposite to each other. Clearly, the cuboids shown in figures (2) and (4) cannot be formed as in each of the two cuboids the two shaded rectangular faces appear adjacent to each other. So, only the cuboids in figures (1) and (3) can be formed.
Question – 3 : A alone can complete the work in 12 days and B alone can complete the work in 24 days. When A, B and C worked together then work will be completed in 7211 days. In how many days C alone can complete the half of the work?
Answer – (A) : 18 days
Answer – (B) : 72 days
Answer – (C) : 36 days
Answer – (D) : 40 days
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Correct Answer : (A)
Explanation : Given:A can do the work in 12 days.B can do the work in 24 days.A, B, C can do the work in 7211 days.A’s 1 days work =112B’s 1 days work =124⇒ (A + B + C)’s 1 days work =112+124+1C⇒112+124+1C=1172⇒1C=136C alone can complete the work in 36 days.∴ Time taken by C to complete half of work =36×12=18 days
Question – 4 : A sum becomes Rs. 8800 after 3 years and Rs. 14872 after 5 years on compound interest. The sum is:
Answer – (A) : Rs. 4005.46
Answer – (B) : Rs. 4005.46
Answer – (C) : Rs. 5207.10
Answer – (D) : Rs. 5307.10
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Correct Answer : (B)
Explanation : Now the formula for amount on compound interest basis can be given asA = P 1 + R100tWhere A = Amount, P = Principal, R = Rate of interest, T = Time periodAs the sum becomes Rs. 8800 after 3 years and 14872 after 5 years the two equations can be written as8800 = P 1+R1003 —— eq. (i)14872 = P 1+R1005 —— eq. (ii)Dividing eq. (ii) by eq. (i)148728800=P1+R1005P1+R1003∴ 1.69 = 1+R1002∴ 1.3 = 1 + R100∴ R100 = 0.3∴ R = 30%Putting the value of R in eq. (i)8800 = P 1+301003∴ 8800 = P × 1.33∴ P = 88001.33∴ P = Rs. 4005.46
Question – 5 : In a class there are total 550 students. Ratio of boys and girls is 6 : 5. How many girls should join the class so that ratio becomes 5 : 6.
Answer – (A) : 110
Answer – (B) : 80
Answer – (C) : 90
Answer – (D) : 110
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Correct Answer : (D)
Explanation : GivenTotal students = 550Ratio of boys and girls = 6 : 5Number of boys = 550 × 611 = 300Number of girls = 550 × 511 = 250Let the number of girls joining the class be x.Now,300(250 + x) = 5 : 6⇒ 250 + x = 300 × 65⇒ 250 + x = 360⇒ x = 360 – 250 = 110∴ The required girls is 110.
Question – 6 : Find the CI earned in 209 years at 18% compounded annually, if the amount invested is Rs. 10000.
Answer – (A) : Rs. 4480.96
Answer – (B) : Rs. 4590.96
Answer – (C) : Rs. 4390.96
Answer – (D) : Rs. 4480.96
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Correct Answer : (D)
Explanation : Given:Amount invested, P = Rs. 10000Time period = 209 years = 1+1+29 yearsFor first year, 100 at a rate of 18% would become, 118⇒ Principal = 100 and Amount = 118Similarly, for second year, 100 at a rate of 18% would become, 118⇒ Principal = 100 and Amount = 118For third year, time is 29 years, hence effective rate would become 4%⇒ 100 at the rate of 4%, would become 104Now principal = Rs. 100000Total Amount = 118 × 118 × 1.04 = Rs. 14480.96⇒ For the principal of Rs. 10000, the total amount would be Rs. 14480.96∴ CI is Rs. 4480.96
Question – 7 : Find the missing number.
Answer – (A) : 9
Answer – (B) : 9
Answer – (C) : 1
Answer – (D) : 4
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Correct Answer : (B)
Explanation : Logic:Central Element = sum of the lower element – top element.9 + 7 – 4 = 124 + 1 – 1 = 4So,x = 6 + 5 – 2 = 9.Missing number = 9.
Question – 8 : The ratio of the present ages of Ranjini and Shahid is 5 ∶ 4. After 13 years the ratio of their ages will be 6 ∶ 5. What is Ranjini’s present age?
Answer – (A) : 65
Answer – (B) : 65
Answer – (C) : 60
Answer – (D) : 32
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Correct Answer : (B)
Explanation : Let present age of Rajini = ALet present age of Sahid is = BThe ratio of present ages of Ranjini and Shahid is 5 : 4⇒ AB=54⇒ 4A = 5B —-(1)After 13 years the ratio of their ages will be 6 : 5.⇒ (A + 13)(B + 13)=65⇒ 5A + 65 = 6B + 78⇒ 5A – 6B = 13 —-(2)From equation (1) and (2)⇒ 554B – 6B = 13⇒ 6.25B – 6B = 13⇒ 0.25B = 13⇒ B = 52From equation 1)⇒ 4A = 5B⇒ A = 65So, present age of Rajini = A = 65 years.
Question – 9 : A can finish a work in ‘x’ days, B in 12 days and C in 18 days. B and C start the work but are forced to leave after 5 days. The remaining work was done by A in 8 days. Then x = ?
Answer – (A) : 26211 days
Answer – (B) : 11 days
Answer – (C) : 26211 days
Answer – (D) : 201125 days
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Correct Answer : (C)
Explanation : A can finish a work in ‘x’ days, B in 12 days and C in 18 days.∴ Part of work A can do in a day =1x∴ Part of work B can do in a day =112∴ Part of work C can do in a day =118Part of work B and C do in 5 days =512+518=15+1036 =2536∴ Remaining part of work =1−2536 =1136The remaining work was done by A in 8 days.∴11361x=8⇒11x=288⇒x=28811=26211 days
Question – 10 : Find 1+12+14+18+116+…+∞= ?
Answer – (A) : 2
Answer – (B) : ∞
Answer – (C) : 2
Answer – (D) : 1
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Correct Answer : (C)
Explanation : Given,1+12+14+18+116+…+∞Common ratio ( r)=121=12,1412=12,11618=12Here, common ratio is same so it’s an example of GP.We know that,Sum of infinite terms of GP∴S∞=a1−rWhere, a = first term, r = common ratio (r ≠ 1)Therefore,In series 1+12+14+18+116+…+∞a=1 , and r=12∴S∞=a1−r⇒S∞=11−12⇒S∞=112⇒S∞=2
Question – 11 : 360, 60, 12, 3, 1, 0.5, ?
Answer – (A) : 0.5
Answer – (B) : 0.25
Answer – (C) : 0.5
Answer – (D) : 0.75
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Correct Answer : (C)
Explanation : The pattern of the given series is as follows:⇒ 360⇒ 60 = 3606⇒ 12 = 605⇒ 3 = 124⇒ 1 = 33⇒ 0.5 = 12⇒ 0.5 = 0.51∴ The missing term in the series = 0.5
Question – 12 : 8, 20, 44, 80, 128, 188, ?
Answer – (A) : 260
Answer – (B) : 248
Answer – (C) : 252
Answer – (D) : 260
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Correct Answer : (D)
Explanation : The given series follows the below pattern:⇒ 20 = 8 + (12 × 1)⇒ 44 = 20 + (12 × 2)⇒ 80 = 44 + (12 × 3)⇒ 128 = 80 + (12 × 4)⇒ 188 = 128 + (12 × 5)The next number in the series would be 188 + (12 × 6)= 188 + 72= 260∴ Required value = 260
Question – 13 : If the under given letters of English alphabet are arranged in a proper sequence, we get the name of a/an _______.[ A A M U Y N]
Answer – (A) : Indian River
Answer – (B) : Indian City
Answer – (C) : Indian Food Item
Answer – (D) : Indian River
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Correct Answer : (D)
Explanation : By rearranging given letters,We get “YAMUNA”.Yamuna is a river.
Question – 14 : Eight members of a club donate Rs. 1000 each towards a Relief Fund and the President of the club donates Rs. 500 more than the average donation of all (including President) of them. Then the contribution of the president is:
Answer – (A) : Rs. 1562.5
Answer – (B) : Rs. 1562.5
Answer – (C) : Rs. 562.5
Answer – (D) : Rs. 2062.5
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Correct Answer : (B)
Explanation : Eight members of a club donate Rs. 1000 eachTotal amount given by 8 members = Rs. 8000Let the average donation of all = Rs. xContribution of the President = Rs. x+500We know that,Average =Sum of observationNo. of observation⇒x=8000+x+5009⇒9x=8500+x⇒8x=8500⇒x=85008∴x= Rs. 1062.5Contribution of the president = Rs. x+500= Rs. 1062.5+500= Rs. 1562.5
Question – 15 : If in a certain code, BAT= 40, CARD = 216, then how will “CAKE” be coded as?
Answer – (A) : 165
Answer – (B) : 165
Answer – (C) : 168
Answer – (D) : 177
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Correct Answer : (B)
Explanation : Alphabets A B C D E F G H I J K L M Positional value 1 2 3 4 5 6 7 8 9 10 11 12 13 Positional value 26 25 24 23 22 21 20 19 18 17 16 15 14 Alphabets Z Y X W V U T S R Q P O N “BAT” has been coded as:B → 2, A → 1, T → 20So, 2 × 1 × 20 → 40“CARD” has been coded as:C → 3, A → 1, R → 18, D → 4So, 3 × 1 × 18 × 4 → 216Similarly, “CAKE” will be coded as:C → 3, A → 1, K → 11, E → 5So, CAKE = 3 × 1 × 11 × 5 = 165Thus, “CAKE” will be coded as “165”.
Question – 16 : In the following question, select the related word from the given alternatives.Butterfly : Lepidopterology : : Map : ?
Answer – (A) : Cartography
Answer – (B) : Phycology
Answer – (C) : Cartography
Answer – (D) : Numismatics
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Correct Answer : (C)
Explanation : The study of butterflies is known as Lepidopterology. Similarly, the study of maps is known as Cartography.
Question – 17 : Every 3 days, we feed our anaconda. Every 14 days, we clean his cage. Today, we cleaned his cage and fed him. How many days from today will we feed him and clean his cage on the same day?
Answer – (A) : 42
Answer – (B) : 42
Answer – (C) : 26
Answer – (D) : 35
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Correct Answer : (B)
Explanation : Multiple and common matching of each number :Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42Multiples of 14 = 14, 28, 42Common Multiple: 42
Question – 18 : By how much is 48 less than 89 ?
Answer – (A) : 718
Answer – (B) : 718
Answer – (C) : 618
Answer – (D) : 719
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Correct Answer : (B)
Explanation : Let the fraction be x.According to the question:⇒ 48 + x = 89⇒ x = 89–12⇒ x = 16–918⇒ x = 718∴ The fraction is 718 .
Question – 19 : Find the missing number. 4820931566?
Answer – (A) : 18
Answer – (B) : 18
Answer – (C) : 16
Answer – (D) : 26
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Correct Answer : (B)
Explanation : 20 = 8 x 2 + 415 = 3 x 2 + 9So,The number on the blank space,6 x 2 + 6 = 18.
Question – 20 : Direction: In a certain code language “Coffee is hot” is written as “yee tee tu”, “Hot soup tasty” is written as “sno tu pnu”, “tasty things coffee” is written as “tee sty pnu”.Then what is the code for “soup is things”.
Answer – (A) : sno yee sty
Answer – (B) : sno yee sty
Answer – (C) : pnu tee sty
Answer – (D) : tee yee sty
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Correct Answer : (B)
Explanation : So, the code forIs = yee;Soup = sno;Things = sty;So, the code for “soup is things” is “sno yee sty”.
Question – 21 : By selling cloth at Rs. 9 per meter, a shopkeeper loses 10% . Find the rate at which it should be sold so as to earn a profit of 15% ?
Answer – (A) : Rs. 11.50
Answer – (B) : Rs. 11.30
Answer – (C) : Rs. 11.40
Answer – (D) : Rs. 11.50
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Correct Answer : (D)
Explanation : According to the question:By selling cloth at Rs. 9 per meter, a shopkeeper loses 10%Therefore, Selling Price =90% of Cost Price⇒ Cost Price =(9×100)90⇒ Cost Price = 10 Rs. per meterTo earn a profit of 15% ,Selling Price =115% of 10⇒ Selling Price =11.50 Rs. per meter
Question – 22 : An isosceles triangle has perimeter 30 cm and each of the equal side is 12 cm. Find the area of the triangle.
Answer – (A) : 915 cm 2
Answer – (B) : 715 cm 2
Answer – (C) : 75 cm 2
Answer – (D) : 915 cm 2
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Correct Answer : (D)
Explanation : Equal sides of the triangle are 12 cm each.Let the third side =x cm.Since, perimeter =30 cm∴12+12+x=30 cm⇒x=30−12−12 cm⇒x=6 cmNow, semi-perimeter =302=15 cm∴ Area of the triangle =s(s−a)(s−b)(s−c)=15(15−12)(15−12)(15−6)=15×3×3×9=5×3×3×3×3×3 cm 2=32×32×3×5=3×3×5×3=915 cm 2Thus, the required area of the triangle =915 cm 2
Question – 23 : If a shirt was bought at 40% discount instead of 30% discount, the customer would have save Rs. 15. Find the selling price of the shirt if it is sold at 50% discount.
Answer – (A) : Rs. 75
Answer – (B) : Rs. 100
Answer – (C) : Rs. 150
Answer – (D) : Rs. 75
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Correct Answer : (D)
Explanation : Given,If a shirt bought 40% discount instead of 30%, then the customer saves Rs. 15.SP = MP × 100-Discount%100Let MP of the shirt be Rs. 100xAccording to the question,100x × 70100 – 100x × 60100 = 15⇒ 70x – 60x = 15⇒ 10x = 15⇒ x = 1510⇒ x = 1.5MP of the shirt = 100 × 1.5 = Rs. 150.∴ Selling price of the shirt if it is sold 50% discount = 150 × 50100 = Rs. 75
Question – 24 : A train speeds past a pole in 15 seconds and speeds past a 100 metre-long platform in 25 seconds. Its length in metres is _____.
Answer – (A) : 150
Answer – (B) : 100
Answer – (C) : 150
Answer – (D) : 200
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Correct Answer : (C)
Explanation : Let l be the length of the train in km and s be its speed in km/hr.∴ While passing the pole,Distance covered = l kmTime = 153600 hrsSpeed = s = l153600 = 240l —(i)While passing the platform,Distance covered = Length of train + Length of platform∴ Distance = (l + 0.1) kmTime = 253600 hrsSpeed = s = l+0.1253600 = 144l + 14.4 —(2)From equation (1) and (2) we get:240l = 144l + 14.4∴ l = 0.15 km∴ The length of the train is 150 m.
Question – 25 : Abhijit sells a book for Rs 880 at 10% profit and another for Rs 270 at 10% loss. His total profit or loss %?
Answer – (A) : 4.54%
Answer – (B) : 4.54%
Answer – (C) : 6%
Answer – (D) : 5%
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Correct Answer : (B)
Explanation : Given:Selling price = Rs. 880And profit = 10%Another Selling price = Rs. 270And loss = 10%Formula:Profit = Cost price – Selling priceProfit% = Profit×100Cost priceCost price = 100100+Profit%×Selling priceCost price = 100100–Loss%×Selling priceAccording to the question:⇒ CP of book = 100100+10%×880⇒ CP = 100110×880⇒ CP = Rs. 800Again,⇒ CP of another = 100100–10%×180⇒ CP = 10090×270⇒ CP = Rs. 300Total CP = (800 + 300) = Rs. 1100Total SP = (880 + 270) = Rs. 1150Profit = (1150 – 1100) = Rs. 50Profit % = 501100 × 100⇒ 5011 = 4.54%∴ Total profit is 4.54%.