Home
Class 14
MATHS
2bar.75+3bar.78=?...

`2bar.75+3bar.78=?`

A

`1bar.03`

B

`1bar.53`

C

`5bar.53`

D

`bar5.54`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem \(2.\overline{75} + 3.\overline{78}\), we first need to understand what the repeating decimals mean. ### Step 1: Convert the repeating decimals to fractions 1. **Convert \(2.\overline{75}\) to a fraction:** Let \(x = 2.\overline{75}\). Then, \(100x = 275.\overline{75}\) (multiplying by 100 shifts the decimal point two places to the right). Subtracting the first equation from the second gives: \[ 100x - x = 275.\overline{75} - 2.\overline{75} \] \[ 99x = 273 \] \[ x = \frac{273}{99} \] Simplifying \(\frac{273}{99}\) gives: \[ x = \frac{91}{33} \] 2. **Convert \(3.\overline{78}\) to a fraction:** Let \(y = 3.\overline{78}\). Then, \(100y = 378.\overline{78}\). Subtracting gives: \[ 100y - y = 378.\overline{78} - 3.\overline{78} \] \[ 99y = 375 \] \[ y = \frac{375}{99} \] Simplifying \(\frac{375}{99}\) gives: \[ y = \frac{125}{33} \] ### Step 2: Add the two fractions Now we need to add \(x\) and \(y\): \[ x + y = \frac{91}{33} + \frac{125}{33} = \frac{91 + 125}{33} = \frac{216}{33} \] ### Step 3: Simplify the result Now, simplify \(\frac{216}{33}\): \[ \frac{216}{33} = \frac{72}{11} \] ### Step 4: Convert back to decimal Now convert \(\frac{72}{11}\) back to a decimal: \[ 72 \div 11 = 6.545454\ldots \] This is \(6.\overline{54}\). ### Final Answer Thus, the final answer is: \[ 2.\overline{75} + 3.\overline{78} = 6.\overline{54} \] ---
Promotional Banner

Topper's Solved these Questions

  • LCM & HCF

    MOTHERS|Exercise MULTIPLE CHOICE QUESTION|200 Videos
  • PARTNERSHIP

    MOTHERS|Exercise MULTIPLE CHOICE QUESTION |51 Videos

Similar Questions

Explore conceptually related problems

if bar a+2 bar b+3bar c=bar 0 then bar axxbar b+bar b xx bar c+bar cxx bara = lambda (bar b xx bar c) ,where lambda=

bar (a) = 2bar (i) + 3bar (j) -bar (k), bar (b) = bar (i) + 2bar (j) -4bar (k), bar (c) = bar (i) + bar (j) + bar (k), bar (d) = bar (i) -bar (j) -bar (k) then

If bar(a)+2bar(b)+3bar(c)=bar(0) then bar(a)xxbar(b)+bar(b)xxbar(c)+bar(c)xxbar(a)=

bar(a)+2bar(b)+3bar(c)=bar(0) and bar(a)xxbar(b)+bar(b)xxbar(c)+bar(c)xxbar(a)=iota(bar(b)xxbar(c)), then iota=

bar(a)+2bar(b)+3bar(c)=bar(0) and bar(a)xxbar(b)+bar(b)xxbar(c)+bar(c)xxbar(a)=rho(bar(b)xxbar(c)), then rho=

The P.V's of the vertices of a triangle are 2bar(i)+3bar(j)+4bar(k),4bar(i)+6bar(j)+3bar(k),3vec i+2bar(j)+3bar(k) P.V.of orthocenter is

Let bar(u)=bar(i)+bar(j),bar(v)=bar(i)-bar(j),bar(w)=bar(i)+2bar(j)+3bar(k) .If bar(n) is a unit vector such that bar(u).bar(n)=0,bar(v).bar(n)=0 then |bar(w).bar(n)|=

Let bar a = 2bar i + bar k,bar b=bar i+bar j+bar k and bar c=4 bar i-3bar j+7 bar k three vectors. The vector which satisfies bar r xx bar b=bar c xx bar b and bar r.bar a=0 is

The p oints O, A, B, C, D are such that bar(OA)= bar a,bar(OB) = bar b,bar(OC) = 2bar a + 3bar b and bar(OD) = bar a- 2 bar b if |bar a|=|3 bar b| then the angle between bar(BD) and bar(AC) is (A) pi (B) pi/2 (C) pi/3 (D) pi/6

bar(b)=bar(i)-2bar(j)-3bar(k),bar(b)=2bar(i)+bar(j)-bar(k),bar(c)=bar(i)+ 3bar(j)-2bar(k) then bar(a).(bar(b)xxbar(c))